Question

A balloon is rising vertically up according to the position equation h(t)=0.2t^2, where h is measured in meters and t in seconds. A camera on the ground is filming the balloon’s ascend 300 meters from the balloons launch site. If x in the angle of elevation of the camera, find the rate of x, in rad/s, at 1 minute after the launch.

Answer 1

To find the rate of the angle of elevation, we need to take the derivative of the position equation with respect to time:

h(t) = 0.2t^2
h'(t) = 0.4t

The angle of elevation of the camera can be found using the tangent function:

tan(x) = h(t) / 300

Taking the derivative of both sides with respect to time t, we get:

sec^2(x) * dx/dt = h'(t) / 300

We want to find the rate of x, dx/dt, at 1 minute after the launch, which is t = 60 seconds. Substituting t = 60 into the position equation, we get:

h(60) = 0.2(60)^2 = 720 meters

Substituting t = 60 and h'(t) = 0.4(60) = 24 into the equation for the angle of elevation, we get:

tan(x) = 720 / 300 = 2.4

Taking the inverse tangent of both sides, we get:

x = tan^-1(2.4) ≈ 67.38 degrees

Substituting x = 67.38 degrees and dx/dt into the equation for the derivative of the tangent function, and converting degrees to radians, we get:

sec^2(67.38°) * dx/dt = 24 / 300
sec^2(1.176 rad) * dx/dt = 0.08
2.214 * dx/dt = 0.08
dx/dt ≈ 0.0361 rad/s

Therefore, the rate of the angle of elevation of the camera at 1 minute after the launch is approximately 0.0361 rad/s.