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Thread: Chapter 4: Applications of Differentiation

  1. #1 Chapter 4: Applications of Differentiation 
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    http://www.thescienceforum.com/Calcu...ions-8994t.php

    Here we are at the end of Chapter 4! I've got an easy one to start things off. My work makes perfect sense to me, but the book's answer doesn't. I am going to use AMax to mean Absolute Max, Amin to mean Absolute min, LMax to mean local max, and Lmin to mean local min.

    Find the local and absoulte extreme values of the function on the given interval.

    5) f(x) = x + sin 2x, [0,π]

    My work:

    f '(x) = 1 + 2 cos 2x

    From what I understand, f(x) can only change from increasing to decreasing if f '(x) = 0 within the given interval. A point at which f '(x) = 0 is a critical point, and only a critical point may be a local min or local max value. To find the criticals point, we set f '(x) = 0 and solve for x.
    f '(x) = 0 = 1 + 2 cos 2x
    cos 2x = -1/2
    2x = -π/3
    x = -π/6

    I would say that since the angle is 2x, that both the cos and sin would completely cycle every π value of x, therefore f '(x) = 0 when x = -π/6, , 2π/6, 5π/6, 8π/6 and so on. 2π/6 and 5π/6 are the only x values within the interval [0,π]
    f(π/3) = Sqrt(3)/2 + π/3
    f(5π/6) = -Sqrt(3)/2 + 5π/6
    f(0) = 1
    f(π) = π

    AMax = π
    LMax = Ø
    Amin = 0
    Lmin = Ø
    (in x values)

    The book answer is:
    AMax = π
    LMax = π/3
    Amin = 0
    Lmin = 2π/3

    Never mind on this one. My problems with this were more or less concerning trig. I think I just need a small break.


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  3. #2  
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    cos(2x) = -1/2 has infinitely many solutions.


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    Yes, I realized that, and I knew that and know that, but I just need a small break.

    I just read about 60 pages of math in the last 8 of my waking hours, all of which was for the most part new material for me.
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  5. #4  
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    Don't blow a gasket!
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  6. #5  
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    Ok, so I've been making pretty good progress. It seems to me, this Chapter 4 is pretty easy in comparison to the other chapters. Anyway, here is a problem that I am stuck on:

    51) Show that the shortest distance from the point (x<sub>1</sub>, y<sub>1</sub>) to the strait line Ax + By + C = 0 is:

    ( | Ax<sub>1</sub> + By<sub>1</sub> + C | ) / Sqrt( A<sup>2</sup> + B<sup>2</sup>)

    I'd show some of the work I've done on it, but I really don't know how I might procede on this one. I recognize that the denominator as the equivilant to what c would be equal to in the pythagorean theorem, but I can't quite figure out how that would fit.

    I guess I have reasoned out that the distance for the line described would be Sqrt( ( | x<sub>1</sub> | - | x | )<sup>2</sup> + ( | y | - | y<sub>1</sub> | )<sup>2</sup>). This formula should only work while both points are in the same quadrant though.
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  7. #6  
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    In general you'll have the distance between your point (x<sub>1</sub>,y<sub>1</sub>) and a point (x,y) on the line is:

    ((x<sub>1</sub>-x)<sup>2</sup> + (y<sub>1</sub>-y)<sup>2</sup>)<sup>1/2</sup>

    Since (x,y) lies on a line, you can get this formula so that the only variable is x. Then you want to find the value of x which makes the distance smallest. I'll let you take it from there.
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    I'm not sure I see how. I know I could say that y = ( -Ax - C ) / B, and that would make the equation:

    ( ( x<sub>1</sub> - x)<sup>2</sup> - (y<sub>1</sub> - ( ( -Ax - C ) / B )<sup>2</sup> )<sup>1/2</sup>

    and while that does eliminate one variable, that doesn't seem to make much progress to eliminating x<sub>1</sub> or y<sub>1</sub> or x. I don't really see much of a relationship between the two points besides knowing the distance between them in terms of three different variables.
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  9. #8  
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    Your point (x<sub>1</sub>, y<sub>1</sub>) is fixed, so its coordinates are constants, not variables. You now have one variable--x.
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    I think maybe my understanding of constants may be wrong. If we complete this problem 51 wouldn't we have an equation that could find the distance between any strait line and any point? Wouldn't that mean the coordinates of the point should be variables since the point is not defined at a particular position?

    I understand π is called a constant because it always represents the number 3.1415...

    I understand e is a constant because it always represents it always represents 2.71828...

    I understand c is a constant for the speed of light, that g is a constant for gravitational acceleration, or maybe m<sub>sun</sub> for the mass of the sun. Why should x<sub>1</sub> and y<sub>1</sub> be constants? I don't understand.
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    You are correct that the end result of this problem is a general formula that will work for any line and any point. But in deriving this formula, you are acting as if you have picked out a specific point and a specific line. When you look at the distance between the specific point and a point on the line, the coordinates of the latter point may vary (there are infinitely many points on the line), whereas the coordinates of the former must stay the same, because they refer to a specific point. Thus, when we're applying the tools of calculus to the distance formula, we do not treat these coordinates as variables. And any symbol representing a quantity in an equation which is not a variable is called a constant in this context.

    So your distance formula is truly a function of 1 variable, x, and you want to minimize it.
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    Ok, I think I understand what you are saying. Help me to understand this though:

    When is a differentiated variable equal to 1, such as I often deal with on the right side of an equation or expression, and when is the differentiated variable equal to something such as y', which I believe would also be called a variable?

    Thinking about what y' means I would say that the symbol represents the rate of change of y. The rate of change may be 1, but is not necisarily. Comparing y and its derivitive y' to another variable that is lets say x and has a derivitive of 1, I would conclude that the rate of change of y may change while the derivitive of x may not. What makes this known?
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  13. #12  
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    A variable's derivative is 1 if it is the variable of differentiation or is set equal to the variable of differentiation. So, differentiating with respect to x, we have x' = 1, and if y = x, then we also have y' = 1.

    If we have some variable y which depends on x in some way, its derivative is written y'. If we know more info, e.g. y = sin x, then we can write y' in terms of x, to, e.g. y' = cos x. If not, the expression y' is as good as we can do.

    If something does not depend on x, then it is treated as a constant when we take derivatives with respect to x. For example, the point (x<sub>0</sub>,y<sub>0</sub>) does not change when we consider different x-coordinates of points on the line Ax+By+C=0, and so the derivative of either coordinate with respect to x is 0.

    I'm not sure what you're asking in your last paragraph. What do you mean by "the rate of change of y may change"?
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  14. #13  
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    This problem was very difficult for me. There were some approaches that I had used, but stopped using because continuing required working out an 81 part polynomial or some enormous amount of work not feasible to do within a few hours time, although I have spent a few days on it now. I decided to use Mathematica to work out a lot of this problem for me. I know almost nothing about the program and am still learning it. I got the idea from a previous poster asking help using it. Anyway, here is my work:

    51)

    The distance between ( x<sub>1</sub>, y<sub>1</sub> ) and the line Ax + By + C I will set equal to D. The equation for D in terms of the variable x is:

    D = ( ( B<sup>2</sup>x<sub>1</sub><sup>2</sup> - 2B<sup>2</sup>xx<sub>1</sub> + B<sup>2</sup>x<sup>2</sup> - 2ABxy<sub>1</sub> - 2BCy<sub>1</sub> + B<sup>2</sup>y<sub>1</sub><sup>2</sup> + A<sup>2</sup>x<sup>2</sup> + C<sup>2</sup> + 2AxC ) / B<sup>2</sup> )<sup>1/2</sup>

    D has the shortest distance when D'=0, so we can find the equation for D':

    D' = ( [ A ( Ax+C) / B<sup>2</sup> ] - x<sub>1</sub> + ( Ay<sub>1</sub> / B ) + x ) / ( [ ( [ Ax + C ] / B ) + y<sub>1</sub> ]<sup>2</sup> + ( x<sub>1</sub> - x )<sup>2</sup> )<sup>1/2</sup>

    we can set D' equal to 0 and solve for x to find the value of x at which D is at its shortest length. Doing so we get the following equation:

    x = (-A<sup>2</sup>x - AC + B<sup>2</sup>x<sub>1</sub> - ABy<sub>1</sub>) / B<sup>2</sup>

    This is an equation that Mathematica worked out for me. I'm not sure why it left an x on the right hand side, but simplifying myself I get:

    x = ( -AC + B<sup>2</sup>x<sub>1</sub> - ABy<sub>1</sub> ) / ( A<sup>2</sup> + B<sup>2</sup> )

    If this is the value of x where D is shortest, then we can plug this into our equation for D in terms of x to get D in terms of constants only where D is at its shortest length:

    D = ( [ ABy<sub>1</sub> + AC + A<sup>2</sup>x<sub>1</sub>]<sup>2</sup> / [ A<sup>2</sup> + B<sup>2</sup> ]<sup>2</sup> + [ ( C / B ) - ( A [ ABy<sub>1</sub> + AC - B<sup>2</sup>x<sub>1</sub> ] / [ B ( A<sup>2</sup> + B<sup>2</sup> ) ] + y<sub>1</sub> ) ]<sup>2</sup> )<sup>1/2</sup>

    I know this is either a very messy problem or a very messy approach. How am I doing so far?

    Actually I think I just found out all of my math I used in Mathematica is wrong because the program thought things like AB were a single variable rather than A*B... That was many hours of work.

    Here is my corrected work:

    D has the shortest distance when D'=0, so we can find the equation for D':

    D' = ( [ A ( Ax+C) / B<sup>2</sup> ] - x<sub>1</sub> + ( Ay<sub>1</sub> / B ) + x ) / ( [ ( [ Ax + C ] / B ) + y<sub>1</sub> ]<sup>2</sup> + ( x<sub>1</sub> - x )<sup>2</sup> )<sup>1/2</sup>

    we can set D' equal to 0 and solve for x to find the value of x at which D is at its shortest length. Doing so we get the following equation:

    x = (-A<sup>2</sup>x - AC + B<sup>2</sup>x<sub>1</sub> - ABy<sub>1</sub>) / B<sup>2</sup>

    This is an equation that Mathematica worked out for me. I'm not sure why it left an x on the right hand side, but simplifying myself I get:

    x = ( -AC + B<sup>2</sup>x<sub>1</sub> - ABy<sub>1</sub> ) / ( A<sup>2</sup> + B<sup>2</sup> )

    If this is the value of x where D is shortest, then we can plug this into our equation for D in terms of x to get D in terms of constants only where D is at its shortest length:

    D = ( [ ( C + Ax<sub>1</sub> + By<sub>1</sub>)<sup>2</sup> ] / [ A<sup>2</sup> + B<sup>2</sup> ] )<sup>1/2</sup>

    or:

    D = ( Ax<sub>1</sub> + By<sub>1</sub> + C) / Sqrt( A<sup>2</sup> + B<sup>2</sup>)

    and this is the answer the book wants! Wow, that took a lot of time.
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  15. #14  
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    So I gather that taking the derivative of the distance formula was the hardest part of this exercise. I apologize, as I should have taken a better look at the task at hand, and there is a trick which makes it much simpler. The thing that makes this derivative so difficult is the square root. If it wasn't there, it'd be much easier. So... we can make it disappear. Why? Well, we're trying to minimize a nonnegative quantity D. If we square D, the minimum still occurs at the same place, as squaring maintains the order of nonnegative quantities. And, since we're only concerned with where the minimum occurs, it doesn't hurt to take the derivative of D<sup>2</sup> instead of D.
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    I needed the practice anyways with deciphering variables, constants, and their derivitives. I did at one point go the D<sup>2</sup> way, but I wasn't sure if it would be faster or not. I think at the time I was more confused about constants and variables and thought that having to divide by 2D would mean that I would have to deal with stuff I wanted to avoid. Anyway, problem 52 was very much like 51, so that wasn't much of a problem. This problem 53 though I've nearly finished, but don't see a way to solve it since I don't see how to get b or r alone:

    53) Find the smallest possible area of an isosceles triangle that is circumscribed about a circle of radius r.

    The diameter of the circle is 2r. The height of the triangle is x + r, where r is the part of the triangle height that extends from the center of the circle to the edge of the circle and x is the remainder of the height. The base of the triangle is equal to b. Using the pathagorean theorem, x = (r<sup>2</sup> - (1/4)b<sup>2</sup>)<sup>1/2</sup>

    Since the value of r does not change, r is a constant, and b is a variable with a domain of (0 , 2π). The area of the triangle in terms of the variable b and the constant r is (b/2)( r + [r<sup>2</sup> - b<sup>2</sup>/4]<sup>1/2</sup>)

    If we want to find the values of b at which A is at a max value or a min value, we first take the derivitive of A which is:

    A'= [ ( r / 2 ) + ( r<sup>2</sup>b / 4 - b<sup>3</sup> / 4 ) ] / [ 2 ( 4r<sup>2</sup>b<sup>2</sup> / 16 - b<sup>4</sup> / 16 )<sup>1/2</sup> ]

    I've simplified this to:

    A' = ( 2r + r<sup>2</sup>b - b<sup>3</sup> ) / ( 2b [ 4r<sup>2</sup> - b<sup>2</sup> ]<sup>1/2</sup> )

    A' is not defined at b = 0 or b = 2r, which does not fall into the domain anyway. From here we set A' = 0 and solve for b in terms of r:

    0 = 2r + r<sup>2</sup>b - b<sup>3</sup>
    b ( b<sup>2</sup> - r<sup>2</sup> ) = 2r

    I dont see any way to seperate b and r. I could factor:

    b ( b + r ) ( b - r ) = 2r

    but that doesn't really help.
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    54) Find the volume of the largest circular cone that can be inscribed in a sphere of radius r.

    I was thinking for this problem that maybe I should assume that the cone is a right circular cone rather than just a circular cone. Would just a circular cone have enough information relating to its sphere that I could break down the area of the cone into a single variable?
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  18. #17  
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    I see a mistake I made on problem 53 while simplifying A'. I'll fix it soon.
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  19. #18  
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    Is there any way you could post a picture of what you're describing in your solution to 53? I'm not sure I understand what you're trying to say.
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    I'm not very good at this picture linking stuff, but here is my try:

    53)



    For convenience:
    53) Find the smallest possible area of an isosceles triangle that is circumscribed about a circle of radius r.

    The diameter of the circle is 2r. The height of the triangle is x + r, where r is the part of the triangle height that extends from the center of the circle to the edge of the circle and x is the remainder of the height. The base of the triangle is equal to b. Using the pathagorean theorem, x = (r<sup>2</sup> - (1/4)b<sup>2</sup>)<sup>1/2</sup>

    Since the value of r does not change, r is a constant, and b is a variable with a domain of (0 , 2π). The area of the triangle in terms of the variable b and the constant r is (b/2)( r + [r<sup>2</sup> - b<sup>2</sup>/4]<sup>1/2</sup>)

    If we want to find the values of b at which A is at a max value or a min value, we first take the derivitive of A which is:

    A'= [ ( r / 2 ) + ( r<sup>2</sup>b / 4 - b<sup>3</sup> / 4 ) ] / [ 2 ( 4r<sup>2</sup>b<sup>2</sup> / 16 - b<sup>4</sup> / 16 )<sup>1/2</sup> ]
    Here is the fix to my mistake in simplifying A':

    A' = r / 2 + ( r<sup>2</sup> - b<sup>2</sup> ) / ( 2 [4r<sup>2</sup> - b<sup>2</sup> ]<sup>1/2</sup> )

    I set A' = 0 to find the values of b at which A is at a min or max value, and simplify:

    3r<sup>4</sup> + r<sup>2</sup>b<sup>2</sup> - b<sup>4</sup> = 0

    I'm not sure where to go from here though.
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    I've used the same techniques in solving problem 54 and I also asumed the cone was a right circular cone.

    r is the constant that describes the radius of the sphere.
    b is base of a two dimensional triangle that extends from the base of the cone to the apex of the cone. b is also the diameter of the base of the cone. r is equal to the part of the height that extends from the center of the sphere to the apex of the cone. x is the part of the triangle height that remains after subracting r. Using the pathagorean theorem, x = (r<sup>2</sup> - b<sup>2</sup>/4)<sup>1/2</sup>. The volume of the cone is represented with V. The equation for the volume of a cone is V = (1/3)bh where b is the base of the cone and h is the height of the cone. Our h = r + x, or using substitution, h = r + (r<sup>2</sup> - b<sup>2</sup>/4)<sup>1/2</sup>. Using substitution on the volume formula for the cone, V = (1/3)b( r + (r<sup>2</sup> - b<sup>2</sup>/4)<sup>1/2</sup>). To find the critical points begin by taking the derivitive of V and get V' = (r/3) + (1/2) [ (4r<sup>2</sup>b<sup>2</sup> - b<sup>4</sup>) / 36 ]<sup>-1/2</sup> [ ( 4r<sup>2</sup>b(2) - 4b<sup>3</sup> ) / 36 ]. I've simplified this to A' = (r/3) + (2r<sup>2</sup> - b<sup>2</sup> ) / ( 3 [ 4r<sup>2</sup> - b<sup>2</sup> ]<sup>1/2</sup> ). Setting A' to 0 and simplifying we get b<sup>2</sup>(b<sup>2</sup> - 4r<sup>2</sup> + 1 ) = 0. The values of b which satisfy this are b = 0, (4r<sup>2</sup> - 1)<sup>1/2</sup>, -(4r<sup>2</sup> - 1)<sup>1/2</sup>. However, the base of a cone can not be negative or 0 so to get the largest area of the cone b = (4r<sup>2</sup> - 1)<sup>1/2</sup>. This answer must be wrong though because when you enter it into the formula of V, you get an imaginary number.
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    This question seems silly:

    58 ) A metal storage tank with volume V is to be constructed in the shape of a right circular cylinder surmounted by a hemisphere. What dimensions will require the least amount of metal?

    My answer would be that obviously a container with an infinitly thin outer shell and a cylinder with an infinitly small height and a heisphere with an infinitely small radius would create a container with the least amount of metal.
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    But then it wouldn't have volume V. I think you are supposed to assume the metal thickness is fixed and differentiate to minimize the surface area.
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    I'll be responding to things in a series of comments. My first comment is that you're misinterpreting the question for number 53. You've drawn an isosceles triangle inscribed in a circle, not one circumscribed about a circle. The former has its vertices lying on the circle and lies inside the circle; the latter has its edges tangent to the circle and lies outside the circle.

    However, we can still attempt to answer a question about inscribed isosceles triangles, but the question would now be what is the largest such triangle. The method of solving this problem would be the same, and since you've set it up, let's try to solve it.

    [[A few minor comments about your setup:

    1. The appropriate domain for b should not be (0,2π) but rather [0,2r]--these are the possibilities for chord lengths in a circle, and any chord can be realized as the base of an inscribed isosceles triangle.

    2. We have to consider both the positive an negative square roots of r<sup>2</sup>-b<sup>2</sup>/4 as values for x, as the height may be smaller than the radius (i.e., when the triangle does not contain the center of the circle). But we can actually discount this case immediately, as any triangle which doesn't contain the center is smaller than the 45-45-90 triangle with b = 2r (as the base and height of such triangles is always smaller). So taking the positive square root is fine.]]

    I just checked your derivative, and I think you need to do it again more carefully, as I got something else. And my answer leads to what I suspected was the solution. We can still solve the equation that your derivative suggests, though...

    b<sup>4</sup>-r<sup>2</sup>b<sup>2</sup>-3r<sup>4</sup>=0

    Use the quadratic formula to solve for b<sup>2</sup>:

    b<sup>2</sup> = (r<sup>2</sup>±(r<sup>4</sup>+12r<sup>4</sup>)/2 = r<sup>2</sup>(1±13<sup>1/2</sup>)/2

    So clearly we want the positive value, so we have:

    b<sup>2</sup> = r<sup>2</sup>(1+13<sup>1/2</sup>)/2
    b = r[(1+13<sup>1/2</sup>)/2]<sup>1/2</sup>

    You can check to see what area this gives you, and then check the area against some "obvious" triangles. You should be able to find one that has area greater than this.
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    Now here's a thought. I think it might be easier to parameterize our triangles not by base length but rather by the angle between the two legs of equal length. This is an angle 0 ≤ θ ≤ π. Let's assume 0 ≤ θ ≤ π/2 (which we deduced in my previous comment), so that your picture is accurate to what I'm about to describe. Geometry shows that the angle made between the two lower radii in your picture is then 2θ, and so the angle made by one of these radii and the altitude is θ. We then have that x = r cos θ and b/2 = r sin θ. The area of our triangle and its derivative with respect to θ are then:

    A = (r+r cos θ)(r sin θ)

    A' = (-r sin θ)(r sin θ)+(r+r cos θ)(r cos θ) = -r<sup>2</sup> sin<sup>2</sup> θ+r<sup>2</sup> cos θ + r<sup>2</sup> cos<sup>2</sup> θ

    Setting this equal to 0, we can cancel out r<sup>2</sup>, and we want to find θ when:

    cos θ + cos<sup>2</sup> θ - sin<sup>2</sup> θ = 0

    Working this out is a good exercise in trigonometric identities, and I don't want to spoil what the answer is anyway.
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    Wow, I made a mistake on the very first step.

    For the inscribed 53)

    I incorrectly took a derivitive of a b<sup>2</sup>. Correcting that error I get:

    A' = (r/2) + (2r<sup>2</sup> - b<sup>2</sup>) / ( 2 [ 4r<sup>2</sup> - b<sup>2</sup> ]<sup>1/2</sup> )

    Setting A' to 0 and solving for b:

    b = r(3)<sup>1/2</sup>

    Taking the area of the triangle with this base:

    A = (3r + r [3]<sup>1/2</sup>) / 2
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  27. #26  
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    Precisely--and the triangle is the equilateral triangle. (Actually, check your area calculation... it should be proportional to r<sup>2</sup>.) The solutions to minimization/maximization problems in geometry tend to be very symmetric--e.g., the perimeter minimizing shape holding a given area is the circle, the surface-area minimizing solid holding a given volume is a sphere, etc., etc., etc., and this jibes with such expectations. If I were more of a geometer, I'd be able to give you better intuition as to why this is true. In any case, we'd also expect the area minimizing circumscribed isosceles triangle to be equilateral. I haven't done the calculations, but it wouldn't be a huge surprise. It'd probably be a good exercise for you to go through that calculation, too. You can set it up analogous to the way you set up the inscribed circles, or you could try something along the lines of what I suggested, parameterizing by angles. You'll have to come up with a nice picture which allows you to play around with geometry, much like the one you drew for this problem.

    Now... let me think about the other problems.
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  28. #27  
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    I agree with Harold regarding 58: don't worry about the thickness of the metal. Just concern yourself with the surface area (which, in reality, is basically directly proportional to the amount of metal, assuming the thickness is predetermined). Also, note that we're given a specific volume V that we want our container to contain, so V should be treated as a constant here. So you should have two equations involving the height h and radius r, one for the volume V, and one for the surface area S:

    V = f(h,r)

    S = g(h,r)

    Since V is fixed, you can use V = f(h,r) to solve for h in terms of r or r in terms of h (one of them is probably easier than the other). Then plug that into S (so you'll be left with S depending on one variable, h or r), and minimize S.
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  29. #28  
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    Okay, this is a lot to digest, so I took a lot of parenthetical comments out and made them footnotes. Ignore them if you wish--they describe slightly more technically some of the intuitive notions I use in what follows.

    For the circular cone, a little thought experiment shows that the answer has to be a right circular cone. Imagine a circular cone inscribed in the sphere. Its volume depends only upon its height h<sup>1</sup> and the radius r of the base, and it grows proportionally to h and to r<sup>2</sup> (exponent, not footnote). We can keep the base fixed and slide the vertex around to create new cones<sup>2</sup>. If the height increases, then the volume necessarily increases. If the altitude is not perpendicular to the circle<sup>3</sup>, then you can push the vertex along the sphere in a manner that makes the altitude higher<sup>4</sup>. So the volume maximizing cone must have its altitude perpendicular to the sphere, and the only way this can happen is if the altitude is actually the line segment joining the vertex and the center of the base, i.e. the cone is right.

    And now let me check your work...

    --------------------------------

    <sup>1</sup> ...as measured via an altitude dropped from the vertex perpendicular to the base... you know this, but the little mathematician in my stomach makes me say these things.

    <sup>2</sup> This is because the sphere is convex, i.e. it contains the line segment between any two points in or on itself, and a cone is really the union of the line segments from the vertex to each point on the base, both of which are assumed to be in/on the sphere.

    <sup>3</sup> ...by which I mean if you can draw a line tangent to the sphere at the vertex which is not perpendicular to the altitude or, equivalently and better yet but possibly more conceptually difficult, the plane tangent to the sphere at the vertex is not perpendicular to the altitude.

    <sup>4</sup> Namely, there must be some direction along the sphere at the vertex (i.e., a tangent line to the sphere at the vertex) which makes an obtuse angle with the altitude, which means it points away from the base, and if you push the vertex in this direction, the altitude will grow.
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  30. #29  
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    Okay, so first things first. The formula for the volume of a cone is:

    V = (area of base)*(height)/3

    You use:

    V = bh/3

    But you use b = "diameter of base". So "area of base" = πb<sup>2</sup>/4, and you should use:

    V = πb<sup>2</sup>h/12

    Intuitively, the cone is a three dimensional object, and so its formula for its volume should grow linearly with an increase in any one dimension. But since the base is circular, the cone grows in two dimensions when the diameter increases, and so the volume grows proportional to the square of the diameter. So, for future reference, this is a good way of checking whether a volume or area formula is "right"--if you're dealing with an n-dimensional object, then the volume should involve a polynomial of total degree n in one-dimensional parameters (variables). If this is gibberish, let me know and I'll try to make it make more sense.
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    This is my diagram for

    circumscribed 53)



    I know this much is right, but if we draw a line that cuts the angle opposite Θ/2, and that line is perpendicular to the line BC, does it pass through the center of the circle? My geometry is rusty.
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  32. #31  
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    Okay, I'm not sure which angle you're talking about, but I'm going to assume you mean the angle at vertex A. The line that you have drawn (which looks to me like it's going through the center of the circle) is not generally perpendicular to BC--in fact, this only happens when you have an equilateral triangle circumscribed about the circle.

    I found a way to make your diagram give enough information for you to work with, but it involved drawing a few more lines--specifically, I connected all of the points on the triangle tangent to the circle. This gave me some more right triangles to work with.

    This is what I came up with on my own, including my thought process that led to my drawing. I need some way of making right angles (because they give me a lot of information). Well, any radius is perpendicular to the circle (or rather, the line tangent to the circle at the point where the radius ends). So using radii is a great idea. So which points on the circle should be the endpoints of my radii? They should also lie on the triangle to help me make some right angles. So how about the three points of the triangle that are tangent to the circle? Using these radii, and extending them a little bit, I got enough info to find the lengths I wanted.
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    Using some of your suggestions, I drew out another diagram, worked out some angle values and use trig to figure out distances. This is where I am at working the problem out:

    circumscribed 53)

    Height = r + r / (sin [Θ/2] )
    Base = 2r(tan [Θ/2] + sec [Θ/2] )
    Area = r<sup>2</sup>( tan [Θ/2] + 2 sec [Θ/2] + sec [Θ/2] csc [Θ/2] )
    A' = r<sup>2</sup>/2( [ 2 ( 1 + sin [Θ/2] ) ] / cos<sup>2</sup> [Θ/2] - 1 / [ sin (Θ/2) ] )

    Setting A' = 0 and simplifying:

    cot<sup>2</sup> [Θ/2] - sin [Θ/2] - 2 = 0

    After deciding to rework the problem again:

    Height = r + r / (sin [Θ/2] )
    Base = 2r tan [Θ/2] / (sin [Θ/2] + 1 )
    Area = r<sup>2</sup>sec [Θ/2]
    A' = ( r<sup>2</sup> / 2 ) ( sec [Θ/2] tan [Θ/2] )

    Setting A' = 0 and simplifying:

    sec [Θ/2] tan [Θ/2] = 0

    Secant never is 0 so we can disreguard it and we are left with:

    tan [Θ/2] = 0
    sin [Θ/2] / cos [Θ/2] = 0
    sin [Θ/2] = 0

    Θ = 0, N(2π) where N is a positive or negative integer.

    However this solution doesn't make sense since it makes the angle Θ too small or large to be in a triangle.
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  34. #33  
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    You know, I did the same calculations (although one of us made a tiny error somewhere, not sure where), but I'm not happy with the answer it gives. I'm going to try it parameterizing by the base, and I'm gonna see what that gives me. I suggest that you give up the angle parameterization, too, because if what we have is right, then it probably involves solving things that, well, I don't know how to solve exactly without looking some shit up.

    Sorry for possibly leading you astray.
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    If my second answer is correct, maybe we could say that the Area of the triangle becomes increasingly small as Θ approaches 0, but that may not make so much sense because that makes the area of the triangle equal to r<sup>2</sup>, the area of the circle.
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  36. #35  
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    Here's the deal... as the angle between the legs of equal sizes approaches 0, the size of the triangle reaches infinity. The triangle approaches an infinite "rectangular" region: a segment of length equal to the diameter tangent to the circle at its midpoint with parallel rays extending from its endpoints also tangent to the circle. Similarly, as the angle goes to π, the triangle approaches a strip: two parallel lines tangent to the circle. So your first answer makes a lot of sense... the derivative has to go off to infinity as it approaches each of these points. Your second answer doesn't satisfy this.
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  37. #36  
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    So I'm bad at taking derivatives, apparently. I just used a graphing utility to find where the derivative of our area function was 0. And, predictably, it's at π/3--the equilateral triangle minimizes the area. I'll see what I can do about doing this by hand...
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  38. #37  
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    I was thinking that about having to be an equallateral triangle since the area of a triangle approaches infinity as Θ gets increasingly large or small, but I'm not sure how to parameterize this triangle using its base and not use Θ.

    Oh, nevermind. I see.
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  39. #38  
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    Okay, I made things work out... I think the both of us were taking derivatives incorrectly. In any case, letting 2t represent the angle at the top of the triangle (so half of it is t), my area was, like you had:

    A = r<sup>2</sup>(tan t + 2 sec t + sec t csc t)

    I factor out sec t:

    A = r<sup>2</sup>(sin t + 2 + csc t)sec t\

    Not really necessary, but for some reason this was the easier way for me to differentiate and work with in general. I got...

    A' = r<sup>2</sup>(cos t - cot t csc t)sec t + (sin t + 2 + csc t)tan t sec t

    Setting equal to 0 and dividing by r<sup>2</sup>:

    0 = (1 - csc<sup>2</sup> t) + (tan<sup>2</sup> t + 2 tan t sec t + sec<sup>2</sup> t)

    And now some heavy duty trig. The Pythagorean theorem implies 1 + cot<sup>2</sup> t = csc<sup>2</sup> t, and the latter parenthetical quantity is factorable:

    0 = -cot<sup>2</sup> t + (tan t + sec t)<sup>2</sup>

    cot<sup>2</sup> t = (tan t + sec t)<sup>2</sup>

    ±cot t = tan t + sec t

    Multiplying by cos t sin t:

    ±cos<sup>2</sup> t = sin<sup>2</sup> t + sin t

    If it's plus on the left, I have:

    cos<sup>2</sup> t - sin<sup>2</sup> t = sin t

    Double angle formula cos 2t = cos<sup>2</sup> t - sin<sup>2</sup> t:

    cos 2t = sin t

    And you can check that this happens only when t = π/6 (for 0 ≤ t ≤ π/2, anyway)<sup>1</sup>. If it's minus on the left, I have:

    -cos<sup>2</sup> t - sin<sup>2</sup> t = sin t

    -1 = sin t

    But this is absurd for 0 ≤ t ≤ π/2. So t = π/6, and since I was letting 2t = π/3 be the angle between the two equal legs, the triangle is equilateral.

    ------------------

    <sup>1</sup> Here's an argument: if cos 2t = sin t, then the triangles with hypotenuse of length 1 and angles t or 2t must be congruent: they share two side lengths (hypotenuse and the side next to 2t or across from t), but they're right triangles, so they must share three sides lengths due to, say, Pythagoras. But then we must have that 2t + t + π/2 = π, or 3t = π/2, so t = π/6.
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    What is wrong with my math on this problem:

    78a) If a projectile is fired with an initial velociy v at an angle of inclination Θ from the horizontal, then its trajectory, neglecting air resistance is the parabola

    y = x tan Θ - x<sup>2</sup>g / ( 2v<sup>2</sup>cos<sup>2</sup>Θ )

    where Θ is greater than or equal to 0 and less than or equal to π/2.

    Suppose the projectile is fired from the base of a plane that is inclined at an angle a, where a > 0, from the horizontal, as shown in the figure. Show that the range of the projectile, measured up the slope, is given by

    R(Θ) = ( 2v<sup>2</sup>cos Θ sin(Θ - a) ) / (g cos<sup>2</sup> a )



    My work:

    We need to find where the line that describes the inclined plane intersects the parabola representing the fired projectile. We already have the equation for the parabola, so the next thing we need is to find the equation for the inclined plane. If the angle at the origin is a, then we can draw a triangle with this angle and the inclined plane of the hypotenuse. If we set the distance of the side adjacent to a equal to x, then taking the tangent of a we find that the distance of the side opposite to a is x tan a. With this we now know two points on the line of inclination: (0.0) and (x, x tan a). We find that the slope between these two lines is tan a, and thus the forumla of the inclined line y = x tan a. To find where these two intersect we can substitute this forumula for the y value in the hyperbola and get:

    x tan a = x tan Θ - x<sup>2</sup>g / (2v<sup>2</sup>cos<sup>2</sup> Θ )
    tan a = tan Θ - xg / (2v<sup>2</sup>cos<sup>2</sup> Θ )
    xg / (2v<sup>2</sup>cos<sup>2</sup> Θ ) = tan Θ - tan a
    x = ( tan Θ - tan a ) ( (2v<sup>2</sup>cos<sup>2</sup> Θ ) / g )
    x = ( sin Θ / cos Θ - sin a / cos a ) ((2v<sup>2</sup>cos<sup>2</sup> Θ ) / g )
    x = 2v<sup>2</sup>sin Θ cos Θ / g - 2v<sup>2</sup>cos<sup>2</sup> Θ sin a / ( g cos a)
    x = ( 2v<sup>2</sup>sin Θ cos Θ cos a - 2v<sup>2</sup> cos<sup>2</sup> Θ sin a ) / ( g cos a )
    x = [ 2v<sup>2</sup> cos Θ( sin Θ cos a - cos Θ sin a ) ] / [g cos a ]
    x = [ 2v<sup>2</sup> cos Θ sin ( Θ - a ) ] / [ g cos a ]

    This value of x should equal the x coordinate of the intersection of the two functions. What we need is the distance of R, so by the pathagorean therom this distance is ( ( [ 2v<sup>2</sup> cos Θ sin ( Θ - a ) ] / [ g cos a ] )<sup>2</sup> + (x tan a)<sup>2</sup> ) <sup>1/2</sup>, but I doubt this will give us the answer the book suggests. I'll work it out though.

    Nevermind, I worked it out and it did give the book answer. Amazing.
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    This problem should be easy. I remember being able to do similar problems very easily years ago, but I can't remember much about how I worked it out.

    75) A canister is dropped from a helicopter 500 meters above the ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of 100 m / s. Will it burst?

    My work:

    I believe that gravitational acceleration is 9.8 m / s<sup>2</sup>, but as of yet I havn't quite figured out an equation that makes sense. I would say that the velocity should be 9.8x, and the distance traveled equal to 9.8x + 9.8(x-1) + 9.8(x-2)...
    There has to be a better way to write this out though. I can say though, if i set v equal to 100, that it will take the canister about 10.2 seconds to reach 100 m / sec, and that if i use the long equation for distance, the distance the canister travels is about 560 m / sec, which does answer my question. No, the canister will not burst if it falls 500 m. It will burst around 560 m though.
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  42. #41  
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    I also am not sure how to do this problem:

    81) Show that, for x > 0

    x / (1+x<sup>2</sup>) < tan<sup>-1</sup>x < x

    My work:

    tan<sup>-1</sup> x = Θ

    In order for me to try to understand the problem I restricted Θ to 0 > Θ > π/2. I then stated that as Θ -> π/2, x -> infinity, and tan Θ -> infinity. I am not sure what to do next though.
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  43. #42  
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    Quote Originally Posted by Demen Tolden
    This problem should be easy. I remember being able to do similar problems very easily years ago, but I can't remember much about how I worked it out.

    75) A canister is dropped from a helicopter 500 meters above the ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of 100 m / s. Will it burst?

    My work:

    I believe that gravitational acceleration is 9.8 m / s<sup>2</sup>, but as of yet I havn't quite figured out an equation that makes sense. I would say that the velocity should be 9.8x, and the distance traveled equal to 9.8x + 9.8(x-1) + 9.8(x-2)...
    There has to be a better way to write this out though. I can say though, if i set v equal to 100, that it will take the canister about 10.2 seconds to reach 100 m / sec, and that if i use the long equation for distance, the distance the canister travels is about 560 m / sec, which does answer my question. No, the canister will not burst if it falls 500 m. It will burst around 560 m though.
    Try v<sup>2</sup> = u<sup>2</sup> + 2as [a standard equation for motion calculation]

    where V= final velocity, u = initial velocity (in this case 0), a = 9.8 and s = distance = 500.

    I get V = 98.99494937m/s

    So it doesn't burst, but only just...

    I get bursting distance as 510.2040816m.

    Hope this helps (and am presuming you're familiar with the standard v=u+at as well.)
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  44. #43  
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    Thanks, but why this equation? I mean why does this equation work?
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  45. #44  
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    Quote Originally Posted by Demen Tolden
    Thanks, but why this equation? I mean why does this equation work?
    It's just derived from the definitions. Dunno if this will help explain, but let's see...

    Acceleration is change in velocity over a period of time. =>

    a = (v - u)/t

    This can be manipulated to give:

    v = u + at (one of the equations I recommended above) Result 1

    or:

    t = (v-u)/a Result 2.

    Now distance travelled is velocity multplied by time =>

    s = ut (for a fixed initial velocity)

    If there is acceleration, then the distance travelled will have to make use of average velocity (assuming constant acceleration):

    s = t * (v+u)/2 Result 3

    Replace the value for t from Result 2 to give:

    s = ((v-u)/a)*((v+u)/2)

    or

    s = (v+u)(v-u)/2a =>

    2as = v<sup>2</sup> - u<sup>2</sup>

    to finally give us:

    v<sup>2</sup> = u<sup>2</sup> + 2as

    Or perhaps this didn't answer your question. The reason 'why' I used that equation was because that was the equation that worked. :-D

    cheer

    shanks
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    No, you did great. Thanks.
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    Sunshinewarrio gave you the right answer but I think he "cheated" a little bit by equating the distance to the average velocity times time. This happens to be correct for a constant acceleration but not necessarily intuitively obvious. If you want to derive it purely from integration you could do the following:
    The final velocity is the acceleration times time
    V=Vo + a t
    distance s=∫ V(t) dt = ∫ Vo+gt dt evaluated from the limits of 0 to final time T
    Integrating we get
    s = Vot + 1/2 gt^2
    The initial velocity is zero and s = height h from which it is dropped, then
    T = sqrt(2h/g)
    substituting for T in the original equation
    V=g sqrt(2h/g) = 9.8 sqrt (2*500/9.8 ) = 98.99 which is less than 100
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    What does it mean to be inversely proportional?

    79) A light is to be placed atop a pole of height h feet to illuminate a busy traffic circle, which has a radius of 40 ft. The intensity of illumination I at any point P on the circle is directly proportional to the cosine of the angle Θ (see the figure) and inversely proportional to the square of the distance d from the source.

    Note: the see the figure note is only a note written in the book. I have not yet uploaded a picture that relates to this problem.
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    Quote Originally Posted by Demen Tolden
    What does it mean to be inversely proportional?

    79) A light is to be placed atop a pole of height h feet to illuminate a busy traffic circle, which has a radius of 40 ft. The intensity of illumination I at any point P on the circle is directly proportional to the cosine of the angle Θ (see the figure) and inversely proportional to the square of the distance d from the source.

    Note: the see the figure note is only a note written in the book. I have not yet uploaded a picture that relates to this problem.
    Gets smaller as the other quantity gets larger.

    Ergo, = n/d<sup>2,</sup>

    where n is a constant.

    Effectively it gets as much smaller as the square of the hypotenuse gets larger. Assuming that the eventual figure will have distance to the point as the hypotenuse for your angle theta.
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  50. #49  
    Forum Bachelors Degree Demen Tolden's Avatar
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    Thanks.

    I think I'm going to move on to the next chapter. There was a problem I still didn't see how to do though.
    81) Show that, for x > 0

    x / (1+x2) < tan-1x < x

    My work:

    tan-1 x = Θ

    In order for me to try to understand the problem I restricted Θ to 0 > Θ > π/2. I then stated that as Θ -> π/2, x -> infinity, and tan Θ -> infinity. I am not sure what to do next though.
    but I think I should be okay. So on to the next chapter,

    Chapter 5: Integrals

    /cheer
    The most important thing I have learned about the internet is that it needs lot more kindness and patience.
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  51. #50  
    Forum Professor sunshinewarrior's Avatar
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    Quote Originally Posted by Demen Tolden
    Thanks.

    I think I'm going to move on to the next chapter. There was a problem I still didn't see how to do though.
    81) Show that, for x > 0

    x / (1+x2) < tan-1x < x

    My work:

    tan-1 x = Θ

    In order for me to try to understand the problem I restricted Θ to 0 > Θ > π/2. I then stated that as Θ -> π/2, x -> infinity, and tan Θ -> infinity. I am not sure what to do next though.
    but I think I should be okay. So on to the next chapter,

    Chapter 5: Integrals

    /cheer
    I saw that one and thought "Trigonometry. Not my scene.", but here are a couple of thoughts:

    Wouldn't it be simpler, instead of restricting Θ and so on, simply to define the relationship the other way:

    x = tanΘ

    Then your problem is proving that:

    tanΘ/(1 + tan<sup>2</sup>Θ) < Θ < tanΘ

    I presume the second part of the inequality would be simple (???!)

    and the first part might be solvable/reducible in terms of further trigonometric identities.

    Anyway, them's my thoughts.

    cheer

    shanks
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  52. #51  
    Forum Professor serpicojr's Avatar
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    Sorry about my lack of participation in this thread for the past few days... The end of the semester gets pretty hectic! I'll try to catch up asap.
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  53. #52  
    Forum Bachelors Degree Demen Tolden's Avatar
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    You don't need to apologize because you aren't always there to answer a question.

    Your help is greatly apreaciated though. See you in the next chapter thread.

    And thanks to Harold and Sunshine for your help too.
    The most important thing I have learned about the internet is that it needs lot more kindness and patience.
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