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An airplane is flying at a speed of 350 mi/h at an altitude of one mile and passes directly over a radar station at time $ t = 0 $.

(a) Express the horizontal distance $ d $ (in miles) that the plane has flown as a function of $ t $ .(b) Express the distance $ s $ between the plane and the radar station as a function of $ d $ .(c) Use composition to express s as a function of $ t $.

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01:52

Clarissa Noh

Calculus 1 / AB

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Functions and Models

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Functions

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Partial Derivatives

Functions of Several Variables

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Okay, so here's the situation. We have the radar station, we have the plane one mile up, and then the plane is flying by. So we're going to call the distance that plane has gone since time zero deep. We want to find the distance as a function of time, and we're told that the rate or the speed is 350 miles per hour and we're used to the common equation. Distance equals rate, times time. So if we substitute the rate in there, we get distance equals 3 50 times T. Now we have the distance s, which is the distance from the radar station to the plane. That's like the high pot news of a right triangle and we want to express s as a function of deep. So let's use the Pythagorean theorem, and we have s squared equals one squared plus D squared. We can square root both sides and we don't need the plus or minus because S is just a length, so we know it's going to be positive. S is the square root of one plus d squared. I simplified one squared. It's just one. So there's ass as a function of D And finally, for part C, we want to find s as a function of time by finding the composition of these two functions we just found. So we're going to put the d function. We're putting the d function inside the S function, so that's going to give us s equals the square root of one plus 3 50 t square. And if we want to simplify that, we can go ahead and find out what 350 square it is. Have a calculator for that. That would be 122,500. So we have s is the square root of one plus 122,500 t.

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