Question

A weather balloon is rising vertically from a launching pad on the ground. A technician standing 300 feet from the launching pad watches the balloon rise. At what rate is the balloon rising at the instant when the angle between the ground and the technician’s line of sight is π 4 radians and is increasing at a rate of 1 120 radian per second?

Answer 1

Answer:

$\dfrac{dh}{dt} =5\ ft/s$

Explanation:

Let

h = height of balloon (in feet).

θ = angle made with line of sight and ground (in radians).

h = 300  tanθ

$\dfrac{dh}{d\theta } = 300 sec^2\theta$

now  $\dfrac{dh}{dt}$ can be written as

$\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}$

$\dfrac{d\theta }{dt} = \dfrac{1}{120}\at \ \theta =\dfrac{\pi}{4}$

When θ = π/4,

$\dfrac{dh}{d\theta } = 300 sec^2\theta$

$\dfrac{dh}{d\theta } = 600$

$\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}$

$\dfrac{dh}{dt} =600\times \dfrac{1}{120}$

$\dfrac{dh}{dt} =5\ ft/s$