Question

A road perpendicular to a highway leads to a farmhouse located 9 mile away. An automobile traveling on the highway passes through this intersection at a speed of 45mph. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 7 miles past the intersection of the highway and the road?

Answer 1

Answer:

27.63 mph

Step-by-step explanation:

First we find the distance, p between the Car and the house using Pythagoras theorem.

$p^2=x^2+y^2\\p^2=7^2+9^2\\p^2=49+81=130\\p=\sqrt{130} miles$

Taking derivatives of:

$p^2=x^2+y^2\\2p \frac{dp}{dt} =2x \frac{dx}{dt} + 2y \frac{dy}{dt}$

Since the farmhouse does not move, its speed $\frac{dy}{dt}=0$

Therefore:

$2*\sqrt{130} *\frac{dp}{dt} =2*7*45 + 2y*0\\2*\sqrt{130} *\frac{dp}{dt}=630\\\frac{dp}{dt}=\frac{630}{2\sqrt{130}}\\\frac{dp}{dt}=27.63 mph$

The distance between the automobile and the farmhouse is increasing at a rate of 27.63mph.

Answer 2

Answer:

27.63 mph

Step-by-step explanation:

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