Re: Angular Rate of Change

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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.

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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.

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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 . Here, so so, the derivative of arctan(5/x) **with respect to x** is . It might be better to multiply both numerator and denominator by to get .

**However**, the crucial point is that , which is what was asked, is .

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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, so . **However**, that is , not ! You have not used the fact that dx/dt= 400 mph.

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

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

Re: Angular Rate of Change

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

Re: Angular Rate of Change

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