Question

A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting directly under the helicopter at a rate of 10 ft/s. Find the rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s.

Answer 1

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is $\sqrt{725}$ ft/ sec

Explanation:

Given:

h(t) =  25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Lets find the derivative of distance with respect to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

Substituting the values of h(t) and  x(t) and simplifying we get,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$  = $\frac{725}{\sqrt{725}}$  = $\sqrt{725}$ ft / sec