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Problem 13 Medium Difficulty

A plane flying horizontally at an altitude of $ 1 mi $ and a speed of $ 500 mi/h $ passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is $ 2 mi $ away from the station.
(a) What quantities are given in the problem?
(b) What is the unknown?
(c) Draw a picture of the situation for any time $ t. $
(d) Write an equation that relates the quantities.
(e) Finish solving the problem.

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November 4, 2019

HORRIBLE video.

William B.

October 11, 2019

I really can't figure out what is being said and written

Jake N.

February 26, 2019

It's so hard to understand.. I get that english isn't your first language but you must get better with writing using PC because it's VERY sloppy and paired with the language barrier is bad.

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Video Transcript

Alright we have an airplane flying overhead and it's flying at 500 mph. And um if from the observer here who uh notices that just at this moment, that diagonal distance away is two miles. And um the vertical hide up is one mile. And our goal is to find basically how fast is the plane moving away from the person. So basically if we call this diagonal distance big D. The top distance acts then we can solve this. So basically um D. Squared is equal to, we're going to do pythagorean theorem. So we'll get X squared plus our +11 mile squared. So that is our D. Squared. We could do square root. But since we're gonna take a derivative it's finally that this way we'll do it implicitly. So we're gonna go ahead and take the derivative because we're dealing with rates. So the derivative of this equation is two times D. D. D. T. Equals two times X. D. X. D. T. And plus zero because the constant derivative is zero. And our goal is to solve for D. D. D. T. The rate of change of distance from the person of the airplane with time. Alright so therefore if we're gonna solve our D. D. D. T. Then we're gonna divide both sides by two. D. C. Two X. D. X. D. T. Over two D. Uh Well um the two's cancel and let's see what values we have for X. We need to solve for it. We can use pythagorean real quick. We're doing it at the moment that the airplane is two miles away. So um X. N. Is the square root of the hypotenuse squared minus the height altitude squared. The X. D. T. We're told is 500 MPH. And then we have over the distance is two miles at the moment. Alright so it looks like we get square root of five square root of four minus one is a square root of three. Um Yes Square root of three. And then 250 if we take 500 divided by two. So this will be 250 Square root of three MPH is the rate of change of the distance from the observer of the airplane with time. So. Alright that was fun. Hopefully that helped. I'll see you next time.

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James Stewart

Calculus: Early Transcendentals

Heather Zimmers

Kristen Karbon

Samuel Hannah

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Heather Zimmers

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Heather Zimmers

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals