Question

A plane flying horizontally at an altitude of 1 mi and a speed of 470 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. (Round your answer to the nearest whole number.)

Answer 1

Answer:

407 mi/h

Step-by-step explanation:

Given:

Speed of plane (s) = 470 mi/h

Height of plane above radar station (h) = 1 mi

Let the distance of plane from station be 'D' at any time 't' and let 'x' be the horizontal distance traveled in time 't' by the plane.

Consider a right angled triangle representing the above scenario.

We can see that, the height 'h' remains fixed as the plane is flying horizontally.

Speed of the plane is nothing but the rate of change of horizontal distance of plane. So, $s=\frac{dx}{dt}=470\ mi/h$

Now, applying Pythagoras theorem to the triangle, we have:

$D^2=h^2+x^2\\\\D^2=1+x^2$

Differentiating with respect to time 't', we get:

$2D\frac{dD}{dt}=0+2x\frac{dx}{dt}\\\\\frac{dD}{dt}=\frac{x}{D}(s)$

Now, when the plane is 2 miles away from radar station, then D = 2 mi

Also, the horizontal distance traveled can be calculated using the value of 'D' in equation (1). This gives,

$2^2=1+x^2\\\\x^2=4-1\\\\x=\sqrt3=1.732\ mi$

Now, plug in all the given values and solve for $\frac{dD}{dt}$. This gives,

$\frac{dD}{dt}=\frac{1.732\times 470}{2}=407.02\approx 407\ mi/h$

Therefore, the distance from the plane to the station is increasing at a rate of 407 mi/h.