# Thread: Related Rates of Change

1. 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 ;

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.

2.

3. 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.

) =

this is in the case of their acute angle,

then,

and i get 0.6394955....

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

4. Originally Posted by Heinsbergrelatz
) =
can you elaborate on how you arrived at this expression?

Originally Posted by Heinsbergrelatz
this is in the case of their acute angle,
As a general rule you should always stick to radians as a measure of angle, particularly in calculus.

5. 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 with respect to the Pythagorean theorem would be .

6. can you elaborate on how you arrived at this expression?
by differentiating the trigonometric ratio of the tangent finction, , but the thing is it could be a related change in

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

7. 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 . 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, should be the equation. Remember to use radians for your angle measure, and that one revolution in radians is equal to radians. Also, why did you only provide one answer for both questions? Or, is the answer for both the same?

8. 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

9. 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.

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

11. Originally Posted by Heinsbergrelatz
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 ;

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.
specifically, what you want to solve is the rate of change in the expression because that is another way to define the line AP.

And, that reduces to:

Which, further reduces to:

And, through a VERY handy identity and a little algebra, we get:

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

12. Originally Posted by Arcane_Mathematician
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 .

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

the answer for a) is
b)
i understand how you derived at
but how come the becomes despite the fact that the answers do align when is inserted. Also the reduction to the formula i quite dont get..

and the miscalculation at the end of your working is due to leaving out the multiplication of which is

wait i think i get it, is it like this, instead of going the long way aroud, its supposed to be: = which then equals ;

anyways thank you for the clarification of the situation.

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.

14. Originally Posted by wallaby
Originally Posted by Arcane_Mathematician
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 .
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.

Originally Posted by Heinsbergrelatz
final answers whited out, simply insert tex tags to view:

the answer for a) is
b)
i understand how you derived at
but how come the becomes despite the fact that the answers do align when is inserted. Also the reduction to the formula i quite dont get..

and the miscalculation at the end of your working is due to leaving out the multiplication of which is

wait i think i get it, is it like this, instead of going the long way aroud, its supposed to be: = which then equals ;

anyways thank you for the clarification of the situation.

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,

15. but it is also valid to do it this way right?;

16. The difference is, where did your come from, and where did the come from? Post all of your work when you can, it will make our lives easier in helping you

17. you see we visualize a triangle, an equilateral triangle due to the AP being 5, thus the radius being 5 and so on. thus all sides are equal. now Bisect the the equilateral traingle, and we can represent the Side AB as x. since we have bisected the triangle in two two equal right angled triangles, the gets divided in to which is the reason why my was divided in two. and then we have a right angled triangle with angles , , and . now we can perform the sine function to find the AP which is represented as

18. okay, then no, that isn't exactly whats happening. That method won't work at, say, . You only get the same answer for this one case, sadly.

19. why wouldn't this method work?, the diameter is fixed in a circle, and the that point "P" happens to coincide a 5cm margin in the circumference from the end of the fixed diameter. that means a fixed radius of 5cm 5cm and 5cm. it makes perfect sense, to both geometrical illustration, and mathematical workings.

well of course it wont work at 45 degrees, because that would change the question completely. 45 degrees would not be an equilateral triangle therefore, it violates my question.

20. right. The problem works in a different way, and just happens to be equivalent to your method at this point. That was the point. It's better to find a method that will give you an answer at any point on the circle, instead of a method that works at only one point. The method that answers your question at every data point will invariably help you understand the workings of the problem better.

21. ah, yes i see that interpretation of yours, yes i agree with you. A way of interpreting the problem that works under all circumstances within a circle related question is indeed more relevant than just one certain viewpoint solution to the problem.

thank you for your help with the problem anyways.

22. Are there any other difficult related rates problems? Please post them if so; I am interested in solving them.

23. alright sure, i have tonnes of them in my book.

for example like this one;

Shaft AB is 30cm long and is attached to a flywheel at A. B is confined to motion along OX. the radius of the wheel rotates clockwise at 100 revolutions per second. Find the rate of change in angle ABO when angle OX is:

a) 120 b) 180 (its in angles not radians)

24. Originally Posted by Heinsbergrelatz
alright sure, i have tonnes of them in my book.

for example like this one;

Shaft AB is 30cm long and is attached to a flywheel at A. B is confined to motion along OX. the radius of the wheel rotates clockwise at 100 revolutions per second. Find the rate of change in angle ABO when angle OX is:

a) 120 b) 180 (its in angles not radians)
this problem needs to be defined a little better. There can't be an angle OX, because we only have 2 points, and we don't know the radius of circle O.

25. you probably cant understand this question fully simply because i cant provided you with a diagram. in the book there is a diagram describing the situation and there surely is AOX. i just thought you guys could visualize alittle ..

26. That was a typo in your previous post then. You wrote "the angle OX", which doesn't make sense.

27. all we need to know is the radius of the circle then.

28. That was a typo in your previous post then. You wrote "the angle OX", which doesn't make sense.
the question in the book states the angle OX. but you can tell its an easy mistake because an angle OX cant exist thereby its an angle AOX if referred to the diagram. (which you guys cant because i cant upload the image up here.)

29. right. And with the radius of the circle we can solve the problem, does the book give that?

30. right. And with the radius of the circle we can solve the problem, does the book give that?
oo yes it does, it is 15cm. sorry my mistake, forgot to state that.

31. Shaft AB is 30cm long and is attached to a flywheel (centered at O) of radius 15cm at A. B is confined to motion along OX. the radius of the wheel rotates clockwise at 100 revolutions per second. Find the rate of change in angle ABO when angle AOX is:

a) b)

The corrected problem of Heinsbergrelatz, so that it needs no diagram.

Method: I will label angle ABO to be and angle AOX to be so;

and so:

a) in radians per second.
b) in radians per second

multiply by for degrees per second.

32. The corrected problem of Heinsbergrelatz, so that it needs no diagram.
apparently though the question you corrected for me was correct, the diagram is already given in the book. Maybe for a better understanding.

but i appreciate your workings to solving the problem, despite the fact i already know how to solve it.

33. corrected only so that people on the forum wouldn't need the problem. I figured you'd know how to solve it already. How goes the studies, how far have you gotten in first semester calc?

34. That one wasn't too difficult; any others?

35. wait arcane, the answers you have provided are not in the nearest degree of accuracy.

a)
b)

im not sure which answers are correct, let me have a go at it, and see if answers collide. (im not saying that anyone is right or wrong here, just that sometimes when your answers with the answer key does not match, you are likely to check it over, or have a double check on it)

That one wasn't too difficult; any others?
did you hit the answers that i provided up there?

corrected only so that people on the forum wouldn't need the problem. I figured you'd know how to solve it already. How goes the studies, how far have you gotten in first semester calc?
thank you for your concerns, yes my studies are going fine. just got the certificate for Physics and chemistry stating an A* for IGCSE standards. im not sure how they do it wherever location you are staying at the moment. calculus bugs me, not as much as statistics, but my scores seem to follow my standards, so ill say im not having extreme difficulties

36. if you tell me how you got that extra derivative of the sine function -derivative of (arcsine) , i can come to an agreeable answer with the answer key. so basically how did that "" come out? thanks

37. Ooops... Yep, a) should have been: \dfrac{400\pi}{\sqrt{13}} my bad. I mistook cosine and sine. b) is, however, correct.

The additional came from the chain rule of differentiation; let then;

38. did you hit the answers that i provided up there?
Yes. Do you have any problems that make greater use of stereometry, though?

39. Yes. Do you have any problems that make greater use of stereometry, though?
alright here is one. Btw what is your level, age?

A and B are two homesteads. A pump house is to be at P on the canal to pump water to both A and B.

a) if A and B are akm and bkm from the canal respectively, show thatkm=kilometres)

b) show that

c) explain why and hence show that

d) hence show that

e) what can be concluded from d)?

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