Question

An observer watches a rocket launch from a distance of 2 kilometers. The angle of elevation, θ is increasing at 3° per second at the instant when θ= π/4. How fast is the rocket climbing at that instant? Make sure you show your work.

Answer 1

At the instant when θ = π/4, the rocket is climbing at a rate of π/15 km/sec.

To find how fast the rocket is climbing at the instant when θ = π/4, we need to use trigonometry and calculus.

Let's define the following variables:

- θ: angle of elevation (in radians)

- h: height of the rocket above the ground (in kilometers)

- t: time (in seconds)

We are given that the angle of elevation, θ, is increasing at 3° per second at the instant when θ = π/4. To convert 3° to radians, we multiply by π/180:

dθ/dt = 3°/sec * π/180 = π/60 rad/sec

We want to find dh/dt, the rate at which the rocket is climbing. Using trigonometry, we know that tan(θ) = h/2. Rearranging the equation, we have h = 2tan(θ).

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

dh/dt = 2 * d(tan(θ))/dt

To find d(tan(θ))/dt, we use the chain rule:

d(tan(θ))/dt = d(tan(θ))/dθ * dθ/dt

The derivative of tan(θ) is sec^2(θ), so we have:

d(tan(θ))/dt =
`sec^2`(θ) * dθ/dt

Substituting this back into the previous equation, we get:

dh/dt = 2 *
`sec^2`(θ) * dθ/dt

At θ = π/4, we have sec(π/4) = √2, so:

dh/dt = 2 *
`(√(2) )^2` * (π/60) = π/15 km/sec