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Related Rates of Change

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Heinsbergrelatz

Posted: Fri Feb 19, 2010 8:32 am Post subject: Related Rates of Change

Forum Ph.D.

Joined: 01 Aug 2009
Posts: 633
Location: Singapore

here is a question im stuck from my revision worksheet, which is not homework. i hope you guys can help me out, thank you in advance.

AOB(O being the center of the circle) is a fixed diameter of a circle of radius 5cm. A point "P" moves around the circle at a constant rate of 1 revolution in 10 seconds. Find the rate at which AP is changing at the instant when:
a)AP is 5cm and increasing
b)P is at B

here is the answer just in case ; $\frac{ \sqrt{3}}{2}\pi$

i can do the previous related rates such as the one involving radian of angles and the related sides, but this one kind of confuses me. Once again thanks alot in advance.

Heinsbergrelatz

Posted: Fri Feb 19, 2010 10:09 am Post subject:

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this is not all but take a look at this question

a rectangle PQRS has PQ of length 20cm and QR increases at a constant rate of 2cm/s. At what rate is the acute angle between the diagonals of the angle changing at the instant when QR is 15cm long?

this is how i solved it; well there are two possible ways.

$tan(\theta) (\dfrac{d\theta}{dt}$ ) = $-20(x)^{-2} (\dfrac{dx}{dt})$

this is in the case of their acute angle, $tan^{-1}(\frac{20}{15})= 53.1$

then,

$(\dfrac{d\theta}{dt} ) = \frac{-20(x)^{-2} (2)}{sec^{2}53.1}$

and i get 0.6394955....

but the problem here is that the answer key says 0.128.... where am i wrong?

Last edited by Heinsbergrelatz on Sat Feb 20, 2010 10:40 pm; edited 1 time in total

wallaby

Posted: Fri Feb 19, 2010 11:51 am Post subject:

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Joined: 03 Jul 2005
Posts: 1386
Location: Australia

Heinsbergrelatz wrote:

$tan(\theta) (\dfrac{d\theta}{dt}$ ) = $-20(x)^{-2} (\dfrac{dx}{dt})$

can you elaborate on how you arrived at this expression?

Heinsbergrelatz wrote:

this is in the case of their acute angle, $tan^{-1}(\frac{20}{15})= 53.1$

As a general rule you should always stick to radians as a measure of angle, particularly in calculus.

Ellatha

Posted: Fri Feb 19, 2010 12:13 pm Post subject:

Forum Junior

Joined: 11 Jul 2009
Posts: 210

Also, fifteen is not a set value, and therefore should be modeled as a variable. Remember also to use the chain rule after you differentiate in problems involving related rates of change, i.e., differentiate implicitly, e.g., the derivative of both sides of c^2 = a^2 + b^2

with respect to the Pythagorean theorem would be $2c\dfrac{dc}{dt} = 2a\dfrac{da}{dt} + 2b\dfrac{db}{dt}$ .

Heinsbergrelatz

Posted: Fri Feb 19, 2010 6:25 pm Post subject:

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Posts: 633
Location: Singapore

Quote:

can you elaborate on how you arrived at this expression?

by differentiating the trigonometric ratio of the tangent finction, $\dfrac{20}{x}$ , but the thing is it could be a related change in $\dfrac{x}{20}$

Quote:

As a general rule you should always stick to radians as a measure of angle, particularly in calculus.

ah yes, i already am aware of that, despite my mistake i made up there.

Before any of this, can you guys help me out with the first one please? thanks

Ellatha

Posted: Sat Feb 20, 2010 8:43 am Post subject:

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Joined: 11 Jul 2009
Posts: 210

For the circle question, it is best to draw a diagram. I believe the idea here is to use the formula for arclength as a measure of distance bewteen points A (fixed) and P (free-moving). The formula is $s = r\theta$ . However, r is not changing here, however the arc length formed bewteen the radii connecting points A and P do, so s and theta are labeled as variables. For your specific case, $s = 5\theta$ should be the equation. Remember to use radians for your angle measure, and that one revolution in radians is equal to $2\pi$ radians. Also, why did you only provide one answer for both questions? Or, is the answer for both the same?

Heinsbergrelatz

Posted: Sat Feb 20, 2010 9:03 am Post subject:

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i believe i gave both the questions a separate answer.
first one was square root 3 multiplied bypi, divided by two, and the second answer is 0.128

wallaby

Posted: Sat Feb 20, 2010 1:20 pm Post subject:

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I think Ellatha is refering to part (b) of the circle question, where the point P is at B. (i'm betting it's zero myself)

I started the circle question by finding the length of the line, connecting A and P, as a function of r and theta. Differentiation of this function will provide you with the time rate of change of |AP|.

As for the rectangle question, try differentiating this instead. $Tan(\frac{\theta}{2})= \frac{20}{x}$

Arcane_Mathematician

Posted: Sat Feb 20, 2010 7:23 pm Post subject:

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you're right wallaby, that's the point where the derivative is 0
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Arcane_Mathematician

Posted: Sat Feb 20, 2010 7:37 pm Post subject: Re: Related Rates of Change

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Heinsbergrelatz wrote:

specifically, what you want to solve is the rate of change in the expression $\sqrt{25(\sin^2\theta+(1-\cos\theta)^2)}$ because that is another way to define the line AP.

And, that reduces to: $5\sqrt{\sin^2\theta+(1-\cos\theta)^2}$

Which, further reduces to: $5\sqrt{2(1-\cos\theta)}$

And, through a VERY handy identity and a little algebra, we get: $10\sin\frac{\theta}{2}$

Heins, as long as you understand how I got there, you're good.

final answers whited out, simply insert tex tags to view:
a) 5\cos\frac{\pi}{6}
b) 5\cos\pi
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wallaby

Posted: Sat Feb 20, 2010 10:35 pm Post subject: Re: Related Rates of Change

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Arcane_Mathematician wrote:

final answers whited out, simply insert tex tags to view:
a) 5\cos\frac{\pi}{6}
b) 5\cos\pi

Your algebra was so much neater than mine, but i think you forgot to multiply by $\frac{d\theta}{dt}$ .

Heinsbergrelatz

Posted: Sun Feb 21, 2010 3:04 am Post subject:

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final answers whited out, simply insert tex tags to view:

the answer for a) is $\frac{\sqrt{3}}{2}\pi$
b)

i understand how you derived at $\sqrt{25(cos\theta)^{2} + (sin\theta)^{2}}$
but how come the $(cos\theta)^{2}$ becomes $(1-cos\theta)^{2}$ despite the fact that the answers do align when $\frac{\pi}{3}$ is inserted. Also the reduction to the formula $10sin\frac{\theta}{2}$ i quite dont get..

and the miscalculation at the end of your working is due to leaving out the multiplication of $\frac{\pi}{5}$ which is $\frac{d\theta}{dt}$

wait i think i get it, is it like this, instead of going the long way aroud, its supposed to be: $sin\frac{\theta}{2}$ = $\frac{0.5x}{5}$ which then equals ;
$10sin\frac{\theta}{2}$

anyways thank you for the clarification of the situation.

Quote:

As for the rectangle question, try differentiating this instead.

thank you for the correction, forgot to half the 90 degree in to 45/45, therefore making it half radian.

Arcane_Mathematician

Posted: Sun Feb 21, 2010 5:14 am Post subject:

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Joined: 21 Feb 2009
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Location: San Jose, California

wallaby wrote:

Arcane_Mathematician wrote:

final answers whited out, simply insert tex tags to view:
a) 5\cos\frac{\pi}{6}
b) 5\cos\pi

Your algebra was so much neater than mine, but i think you forgot to multiply by $\frac{d\theta}{dt}$ .

I left out the the derivative of theta to make sure he was paying attention Wink

Plus, I figure it's important for people to check my work instead of just accept it, so in this I wanted to leave out a step for the person I'm helping to correct. imo, it shows how well he understands and is paying attention.

Heinsbergrelatz wrote:

final answers whited out, simply insert tex tags to view:

the answer for a) is $\frac{\sqrt{3}}{2}\pi$
b)

Quote:

As for the rectangle question, try differentiating this instead.

thank you for the correction, forgot to half the 90 degree in to 45/45, therefore making it half radian.

1-\cos\theta=\versin\theta and wikipedia shows why it's useful. Essentially, all you care about are sine theta, because it gives you one leg of a right triangle on which AP lies, and versine(1-cos) theta gives you the other leg to that triangle. It's in the main picture on the link, the green line that extends from the altitude to the circle.

As for the reduction, it's the half angle formula of sine, also in the link, $\displaystyle\sin \frac{\theta}{2} = \pm\, \sqrt{\frac{1 - \cos \theta}{2}}$
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Heinsbergrelatz

Posted: Sun Feb 21, 2010 5:47 am Post subject:

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Joined: 01 Aug 2009
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but it is also valid to do it this way right?;

$sin\frac{\theta}{2} = \frac{\frac{x}{2}}{5}$

Arcane_Mathematician

Posted: Sun Feb 21, 2010 1:56 pm Post subject:

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Joined: 21 Feb 2009
Posts: 2571
Location: San Jose, California

The difference is, where did your $sin\frac{\theta}{2}$ come from, and where did the $\dfrac{\frac{x}{2}}{5}$ come from? Post all of your work when you can, it will make our lives easier in helping you
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