# Angular Rate of Change

• Nov 7th 2013, 09:13 AM
nycmath
Angular Rate of Change
An plane flies at a height of 5 miles to a point directly over an observer. Consider theta and x. Let x be the distance from the observer to the point directly below the plane. I made a picture of a right triangle.

The question tells me to write theta as a function of x. So, my answer is theta = arctan(5miles/x). Is this correct?

Part B: The speed of the plane is 400 mph. Find d(theta)/dt when x is 10 miles and when x is 3 miles.

The answer must be in radians because I am searching for angle theta. Is this right?

I found d(theta)/dt to be -5/x^2 divided by 1 + (25/x^2). Is this the correct function? It is a complex fraction.

Afterward, I simply plugged x = 10 and x = 3 individually and simplified.

For x = 10, my answer is -1/25.
For x = 3, my answer is -5/34.

Is this right?

This question is from my single variable calculus text, chapter on finding the derivative of inverse functions. Thanks....
• Nov 7th 2013, 09:43 AM
HallsofIvy
Re: Angular Rate of Change
Quote:

Originally Posted by nycmath
An plane flies at a height of 5 miles to a point directly over an observer. Consider theta and x. Let x be the distance from the observer to the point directly below the plane. I made a picture of a right triangle.

The question tells me to write theta as a function of x. So, my answer is theta = arctan(5miles/x). Is this correct?

Yes, it is. Note that theta= arccot(x/5) is also correct.

Quote:

Part B: The speed of the plane is 400 mph. Find d(theta)/dt when x is 10 miles and when x is 3 miles.

The answer must be in radians because I am searching for angle theta. Is this right?
No, the answer must be in radians per hour (or in terms of an angle measure over a time measure) because you area asked for the rate of change of theta.

Quote:

I found d(theta)/dt to be -5/x^2 divided by 1 + (25/x^2). Is this the correct function? It is a complex fraction.
No, that is not correct. You have missed one important point.

You have correctly written theta= arctan(5/x). The derivative of arctan(u) is $\dfrac{1}{1+u^2}\dfrac{du}{dx}$. Here, $u= 5/x= 5x^{-1}$ so $du/dx= -5x^{-2}= -5/x^2$ so, the derivative of arctan(5/x) with respect to x is $\dfrac{-5/x^2}{1+ 25/x^2}$. It might be better to multiply both numerator and denominator by $x^2$ to get $\dfrac{-5}{x^2+ 25}$.

However, the crucial point is that $d\theta/dt$, which is what was asked, is $\dfrac{-5}{x^2+ 25}\dfrac{dx}{dt}$.

Quote:

Afterward, I simply plugged x = 10 and x = 3 individually and simplified.

For x = 10, my answer is -1/25.
For x = 3, my answer is -5/34.
If x= 10, $x^2+ 25= 125$ so $\dfrac{-5}{x^2+ 1}= \dfrac{-5}{125}= -\dfrac{1}{25}$. However, that is $d\theta/dx$, not $d\theta/dt$! You have not used the fact that dx/dt= 400 mph.

Quote:

Is this right?
No. It isn't. You did all of the calculations correctly but you did not answer the question that was asked!

Quote:

This question is from my single variable calculus text, chapter on finding the derivative of inverse functions. Thanks....
• Nov 9th 2013, 05:48 AM
nycmath
Re: Angular Rate of Change
I truly appreciate your help and guidance with this question that is not explained in my text.
• Nov 10th 2013, 02:38 AM