Question 1127029
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We need to find the relationship between the change in the angle of elevation and the speed of the plane -- i.e., the change in the plane's distance from the tracking station.<br>
Let x be the (horizontal) distance of the plane from the ground station; let y be the angle of elevation.  Then, since the plane is flying at an altitude of 3km, x and y are related by<br>
{{{tan(y) = 3/x}}}<br>
Find the derivative with respect to time and solve for the plane's speed, dx/dt.<br>
{{{sec^2(y)(dy/dt) = (-3/x^2)(dx/dt)}}}<br>
{{{dx/dt = (-x^2/3)sec^2(y)(dy/dt)}}}<br>
The given information is that when the angle of elevation is pi/4, the rate of change in the angle of elevation is -pi/4 rad/min.  At an angle of pi/4, x is 3, cos(x) is sqrt(2)/2, cos^2(x) = 1/2, sec^2(x) = 2.  So<br>
{{{dx/dt = (-9/3)(2)(-pi/4) = 3pi/2}}}<br>
The plane's speed is 3pi/2 km/min, or about 283 km/hr, about 177 mph.