# Simple Related Rates problem

• Nov 6th 2010, 08:53 PM
Vamz
Simple Related Rates problem
http://img151.imageshack.us/img151/7...01106at104.png

This is a problem from my practice midterm. No answers are given. Can someone please tell me if I have answered this problem correctly?

S=Distance between P & Airplane

$\displaystyle \displaystyle sin \theta=\frac{5000}{S}$

I take the derivative of $\displaystyle \theta$ & S with respect to $\displaystyle \theta$

$\displaystyle \displaystyle cos \theta=\frac{5000}{\frac{dS}{d\theta}}$

$\displaystyle \displaystyle \frac{dS}{d\theta}=\frac{5000}{cos\theta}=5773.5$

Is my work correct?
Is it OK that I only took the derivative of S (with respect to theta) on the denominator only on the right hand side, or should I have taken the derivative of the entire right hand side using the quotient rule?

In my practice problems, when doing implicit derivatives, we take the derivatives of both sides in their entirety. However, I am watching a youtube video on related rates regarding the volume of a balloon, and the guy only takes the derivatives of the two variables he is interested in. Can someone please elaborate on this?

Thank you!
• Nov 7th 2010, 05:14 AM
Quote:

Originally Posted by Vamz
http://img151.imageshack.us/img151/7...01106at104.png

This is a problem from my practice midterm. No answers are given. Can someone please tell me if I have answered this problem correctly?

S=Distance between P & Airplane

$\displaystyle \displaystyle sin \theta=\frac{5000}{S}$

I take the derivative of $\displaystyle \theta$ & S with respect to $\displaystyle \theta$

$\displaystyle \displaystyle cos \theta=\frac{5000}{\frac{dS}{d\theta}}$

$\displaystyle \displaystyle \frac{dS}{d\theta}=\frac{5000}{cos\theta}=5773.5$

Is my work correct?
Is it OK that I only took the derivative of S (with respect to theta) on the denominator only on the right hand side, or should I have taken the derivative of the entire right hand side using the quotient rule?

In my practice problems, when doing implicit derivatives, we take the derivatives of both sides in their entirety. However, I am watching a youtube video on related rates regarding the volume of a balloon, and the guy only takes the derivatives of the two variables he is interested in. Can someone please elaborate on this?

Thank you!

From the question, the "rate" is wrt the angle rather than time as no speed is given.

Yes, you need a correct method to take the derivative..

You are asked to find $\displaystyle \displaystyle\frac{dS}{d\theta}$

$\displaystyle \displaystyle\ S=\frac{5000}{Sin\theta}$

$\displaystyle \displaystyle\frac{dS}{d\theta}=\frac{d}{d\theta}\ left(\frac{5000}{Sin\theta}\right)=5000\frac{d}{d\ theta}\left(\frac{1}{Sin\theta}\right)=5000\frac{d }{d\theta}Cosec\theta}=5000\frac{d}{d\theta}\left( Sin\theta\right)^{-1}$
• Nov 7th 2010, 07:59 AM
Vamz
Thanks. I'm not quite sure I understand what you did here. First of all, you solved for "S".
I thought with implicit differentiation, you should not do this, and solve for dy/dx at the very end?

Ok, then you differentiated the left hand side. Makes sense. What exactly are you doing to the right hand side? This isnt the quotient rule is it?

Are we not taking the deriv of the ENTIRE side, but JUST the variables we are interested in, correct? In this case it was $\displaystyle \frac{1}{sin\theta}=cosec\theta$?

Why couldnt you have just done: $\displaystyle \frac{5000}{\frac{d}{d\theta}sin\theta}=\frac{5000 }{cos\theta}$?
• Nov 7th 2010, 08:07 AM
skeeter
Quote:

Originally Posted by Vamz
Thanks. I'm not quite sure I understand what you did here. First of all, you solved for "S".
I thought with implicit differentiation, you should not do this, and solve for dy/dx at the very end?

Ok, then you differentiated the left hand side. Makes sense. What exactly are you doing to the right hand side? This isnt the quotient rule is it?

Are we not taking the deriv of the ENTIRE side, but JUST the variables we are interested in, correct? In this case it was $\displaystyle \frac{1}{sin\theta}=cosec\theta$?

Why couldnt you have just done: $\displaystyle \frac{5000}{\frac{d}{d\theta}sin\theta}=\frac{5000 }{cos\theta}$?

you're taking the derivative of a function w/respect to a variable, specifically, the function $\displaystyle S(\theta)$ w/r to $\displaystyle \theta$.

the derivative of $\displaystyle csc{\theta} = \frac{1}{\sin{\theta}}$ is not $\displaystyle \frac{1}{\cos{\theta}} = \sec{\theta}$
• Nov 7th 2010, 08:14 AM
Quote:

Originally Posted by Vamz
Thanks. I'm not quite sure I understand what you did here. First of all, you solved for "S".
I thought with implicit differentiation, you should not do this, and solve for dy/dx at the very end?

Ok, then you differentiated the left hand side. Makes sense. What exactly are you doing to the right hand side? This isnt the quotient rule is it?

Are we not taking the deriv of the ENTIRE side, but JUST the variables we are interested in, correct? In this case it was $\displaystyle \frac{1}{sin\theta}=cosec\theta$?

Why couldnt you have just done: $\displaystyle \frac{5000}{\frac{d}{d\theta}sin\theta}=\frac{5000 }{cos\theta}$?

By doing it that way, you are not correctly differentiating the right-hand-side.

The question asks "At what rate is the distance s between the airplane and the fixed point P changing with $\displaystyle \theta$ when $\displaystyle \theta=30^o$."

This is asking for $\displaystyle \displaystyle\frac{ds}{d\theta}$ evaluated for $\displaystyle \theta=30^o$

Hence, we write $\displaystyle s$ in terms of $\displaystyle \theta$ and differentiate.

One option is to use the quotient rule for "either" of the following equivalent expressions.

$\displaystyle \displaystyle\frac{ds}{d\theta}=\frac{d}{d\theta}\ left(\frac{5000}{Sin\theta}\right)=5000\frac{d}{d\ theta}\left(\frac{1}{Sin\theta}\right)$

Alternatively, you can differentiate one of the alternative expressions.

But you must differentiate the entire expression of the RHS, not just the denominator.
• Nov 7th 2010, 09:34 AM
Vamz
Ok, I think I follow you.

so we have
$\displaystyle \displaystyle\frac{ds}{d\theta}=RHS$

and we have to take the derivative of the entire RHS. But, since we are only interested in the $\displaystyle Sin\theta$
part of it, we are allowed to pull the 500 outside the derivative expression.

$\displaystyle \frac{d}{d\theta}\left(\frac{5000}{Sin\theta}\righ t)=5000\frac{d}{d\theta}\left(\frac{1}{Sin\theta}\ right)$

So, now we are allowed to treat $\displaystyle \frac{1}{sin\theta}=(sin\theta)^{-1} OR = csc\theta$ as one term which we can take the derivative of, correct?

You stopped at:
$\displaystyle 5000\frac{d}{d\theta}Cosec\theta}=5000\frac{d}{d\t heta}\left(Sin\theta\right)^{-1}$

Quote:

Originally Posted by skeeter
you're taking the derivative of a function w/respect to a variable, specifically, the function $\displaystyle S(\theta)$ w/r to $\displaystyle \theta$.

the derivative of $\displaystyle csc{\theta} = \frac{1}{\sin{\theta}}$ is not $\displaystyle \frac{1}{\cos{\theta}} = \sec{\theta}$

I thought the derivative of $\displaystyle csc\theta = -csc\theta*cot\theta$ ?

But,
couldnt you go further?
$\displaystyle 5000\frac{d}{d\theta}Cosec\theta}=5000(-csc\theta*cot\theta)$

$\displaystyle \frac{ds}{d\theta}=5000(-csc\theta*cot\theta)$
• Nov 7th 2010, 09:39 AM
skeeter

$\displaystyle S = 5000\csc{\theta}$

$\displaystyle \displaystyle\frac{dS}{d\theta} = -5000\csc{\theta}\cot{\theta}$
• Nov 7th 2010, 09:47 AM
When skeeter wrote....

"the derivative of $\displaystyle cosec\theta=\frac{1}{sin\theta}$ is not $\displaystyle \frac{1}{cos\theta}=sec\theta$"

he meant $\displaystyle cosec\theta=\frac{1}{sin\theta}$ and $\displaystyle \frac{1}{cos\theta}=sec\theta$

and the derivative of $\displaystyle cosec\theta$ is not $\displaystyle sec\theta$
• Nov 7th 2010, 10:03 AM
Vamz
IC. So finally, when I plug in 30, I get -17320.50ft/degree for a final answer.

I also tried using the quotient method for the RHS
$\displaystyle \frac{d}{d\theta}(\frac{5000}{sin\theta})=\frac{-5000*cos\theta}{sin^2\theta}$

When I plug in 30, this also gives me -17320.50ft/degree

So I guess, both methods are perfectly acceptable right?
• Nov 7th 2010, 02:43 PM
skeeter
Quote:

Originally Posted by Vamz
IC. So finally, when I plug in 30, I get -17320.50ft/degree for a final answer.

I also tried using the quotient method for the RHS
$\displaystyle \frac{d}{d\theta}(\frac{5000}{sin\theta})=\frac{-5000*cos\theta}{sin^2\theta}$

When I plug in 30, this also gives me -17320.50ft/degree

So I guess, both methods are perfectly acceptable right?

yes ... however, (and I know this will open a can of worms) , the units are -17320.50 ft/radian = -302.3 ft/degree
• Nov 8th 2010, 10:18 AM
As a guide, while the elevation of the plane increases by $\displaystyle 1^o$ viewed from P,
the distance $\displaystyle s$ from P to the plane only decreases by a small fraction.
The average rate of change of $\displaystyle s$ at around $\displaystyle 30^o$ would be approximately
$\displaystyle \displaystyle\ s_2-s_1=5000\left(\frac{1}{sin30.5^o}-\frac{1}{sin29.5^o}\right)=9,851.5-10,153.9=-302.4\;\ ft/^o$