# Related rate

## Homework Statement

A winch at the top of 12 meter high building pulls a pipe of the same length to a vertical position. The winch pulls in rope at a rate of -0.2 meters per second. Find the rate of the vertical change and the rate of the horizontal change at the end of the pipe when y=6.

## Homework Equations

http://www.imagehosting.com/out.php/i249658_ahhh.JPG

## The Attempt at a Solution

I simply don't have enough information..

I set up an equation relating the three variables..

$$s^{2}=x^{2}+(12-y)^{2}$$

and differentiate with respect to time and end up with

$$s\frac{ds}{dt}=x\frac{dx}{dt}-(12-y)\frac{dy}{dt}$$

But I simply don't have all of the variables needed. I can solve for x (with pythagorean theorem), but not dx/dt...

Any help is appreciated. :yuck:

## Answers and Replies

arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
Hint:
First find the ANGULAR rate of change by a smart law of cosines.
THEN, find the rates of change of (x,y) by using a right-angled triangle.

orangeflover
Thankyou arildno

I was able to get the correct numerical answer but something is bothering me...

Take the angle made at the top of the wall to be $$\theta$$...

We know that $$\frac{d\theta}{dt}$$ must be positive since the angle is tending towards 90deg right?

I got $$\frac{d\theta}{dt}=\frac{/sqrt{3}}{80}$$... but this leads to something that is incorrect. Because when I set up the following equation to relate x and s, I get this..

$$sin\theta=\frac{x}{s} = x(s)^{-1}$$
$$cos\theta\frac{d\theta}{dt}=x(-1)s^{-2}\frac{ds}{dt}+s^{-1}\frac{dx}{dt}$$

which eventually leads to

$$\frac{dx}{dt}=s(cos\theta\frac{d\theta}{dt}+x*s^{-2}\frac{ds}{dt})$$

Now, solving for x using the right triangle at the base, $$x=\sqrt{12^{2}-6^{2}} = 6*\sqrt{3}$$

and s=12 by similiar reasoning

Then $$tan\theta=\frac{6\sqrt{3}}{6}$$

$$\theta=\frac{\pi}{3}$$

So solving $$\frac{dx}{dt}=s(cos\theta\frac{d\theta}{dt}+x*s^{-2}\frac{ds}{dt})$$

I get $$\frac{dx}{dt}=\frac{\sqrt{3}}{15}$$

Which is the correct magnitude, but the incorrect sign... :/

Last edited:
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
Well, I used the lowest angle, with the variable length s(t) as the opposite side. I didn't calculate it through, but my signs looked good.

Ok, let me try that and see if I can get it to work.

Wow it worked very smoothly using the lower angle... Correct sign and all. I'll have to think about this problem for a bit. Thanks again.
The text passage is a discussion of a related rates problem in calculus, where a student is trying to solve for the rates of change of the horizontal and vertical positions of a pipe being pulled vertically by a winch. The student has set up the problem using the Pythagorean theorem and is having trouble solving for the missing variables. Another user provides a hint to use the law of cosines to find the angular rate of change, which the student then uses to solve the problem.
single_question_with_answer

the passage contains a single mathematical question (how to find the rates of change of the horizontal and vertical positions of the pipe) and an answer is provided in the form of a hint and the student's subsequent solution.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Related rate

## Homework Statement

A winch at the top of 12 meter high building pulls a pipe of the same length to a vertical position. The winch pulls in rope at a rate of -0.2 meters per second. Find the rate of the vertical change and the rate of the horizontal change at the end of the pipe when y=6.

## Homework Equations

http://www.imagehosting.com/out.php/i249658_ahhh.JPG

## The Attempt at a Solution

I simply don't have enough information..

I set up an equation relating the three variables..

$$s^{2}=x^{2}+(12-y)^{2}$$

and differentiate with respect to time and end up with

$$s\frac{ds}{dt}=x\frac{dx}{dt}-(12-y)\frac{dy}{dt}$$

But I simply don't have all of the variables needed. I can solve for x (with pythagorean theorem), but not dx/dt...

Any help is appreciated. :yuck:

## Answers and Replies

arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
Hint:
First find the ANGULAR rate of change by a smart law of cosines.
THEN, find the rates of change of (x,y) by using a right-angled triangle.

orangeflover
Thankyou arildno

I was able to get the correct numerical answer but something is bothering me...

Take the angle made at the top of the wall to be $$\theta$$...

We know that $$\frac{d\theta}{dt}$$ must be positive since the angle is tending towards 90deg right?

I got $$\frac{d\theta}{dt}=\frac{/sqrt{3}}{80}$$... but this leads to something that is incorrect. Because when I set up the following equation to relate x and s, I get this..

$$sin\theta=\frac{x}{s} = x(s)^{-1}$$
$$cos\theta\frac{d\theta}{dt}=x(-1)s^{-2}\frac{ds}{dt}+s^{-1}\frac{dx}{dt}$$

which eventually leads to

$$\frac{dx}{dt}=s(cos\theta\frac{d\theta}{dt}+x*s^{-2}\frac{ds}{dt})$$

Now, solving for x using the right triangle at the base, $$x=\sqrt{12^{2}-6^{2}} = 6*\sqrt{3}$$

and s=12 by similiar reasoning

Then $$tan\theta=\frac{6\sqrt{3}}{6}$$

$$\theta=\frac{\pi}{3}$$

So solving $$\frac{dx}{dt}=s(cos\theta\frac{d\theta}{dt}+x*s^{-2}\frac{ds}{dt})$$

I get $$\frac{dx}{dt}=\frac{\sqrt{3}}{15}$$

Which is the correct magnitude, but the incorrect sign... :/

Last edited:
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
Well, I used the lowest angle, with the variable length s(t) as the opposite side. I didn't calculate it through, but my signs looked good.

Ok, let me try that and see if I can get it to work.

Wow it worked very smoothly using the lower angle... Correct sign and all. I'll have to think about this problem for a bit. Thanks again.
user
The text passage is a discussion of a related rates problem in calculus, where a student is trying to solve for the rates of change of the horizontal and vertical positions of a pipe being pulled vertically by a winch. The student has set up the problem using the Pythagorean theorem and is having trouble solving for the missing variables. Another user provides a hint to use the law of cosines to find the angular rate of change, which the student then uses to solve the problem.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage contains a single mathematical question (how to find the rates of change of the horizontal and vertical positions of the pipe) and an answer is provided in the form of a hint and the student's subsequent solution.