# Thread: Angular Rate of Change

1. ## 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....

2. ## Re: Angular Rate of Change

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.

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.

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 $\dfrac{-5}{x^2+ 25}\dfrac{dx}{dt}$.

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.

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

This question is from my single variable calculus text, chapter on finding the derivative of inverse functions. Thanks....

3. ## Re: Angular Rate of Change

I truly appreciate your help and guidance with this question that is not explained in my text.

4. ## Re: Angular Rate of Change

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