A plane flying horizontally at an altitude of $1 \mathrm { mi }$ and a speed of $550 \mathrm { mi } / \mathrm { h }$ 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.
Select the correct answer.
A) $\approx 476 \mathrm { mi } / \mathrm { h }$
B) $\approx 670 \mathrm { mi } / \mathrm { h }$
C) $\approx 455 \mathrm { mi } \mathrm { h }$
D) $\approx 570 \mathrm { mi } / \mathrm { h }$
E) $\approx 495 \mathrm { mi } / \mathrm { h }$

Question

Quizplus · Accepted Answer

The correct answer is A.Explanation:The distance from the plane to the station is given by the Pythagorean Theorem:$$d=\sqrt{1^2+x^2}$$where x is the horizontal distance from the plane to the station.Differentiating both sides with respect to time, we get:$$\frac{d}{dt}\sqrt{1^2+x^2}=\frac{d}{dt}\sqrt{x^2}$$$$\frac{1}{2\sqrt{1^2+x^2}}\frac{d}{dt}(1^2+x^2)=\frac{1}{2\sqrt{x^2}}\frac{d}{dt}x^2$$$$\frac{1}{2\sqrt{1^2+x^2}}(2x\frac{dx}{dt})=\frac{1}{2\sqrt{x^2}}(2x\frac{dx}{dt})$$$$\frac{x}{\sqrt{1^2+x^2}}\frac{dx}{dt}=\frac{x}{\sqrt{x^2}}\frac{dx}{dt}$$$$\frac{x}{\sqrt{1^2+x^2}}=\frac{x}{\sqrt{x^2}}$$$$\sqrt{1^2+x^2}=\sqrt{x^2}$$$$1^2+x^2=x^2$$$$1^2=0$$This is a contradiction, so there is no value of x that satisfies the equation.

A Plane Flying Horizontally at an Altitude Of 1mi1 \mathrm { mi }1mi

A Plane Flying Horizontally at an Altitude Of $1 \mathrm { mi }$