3. a person is watching the space shuttle launch. the person is 3000 ft from the launch pad. how fast is the distance between the person and the shuttle changing when the shuttle is 4,000 ft high and rising at a rate of 800 ft/sec?



Answer :

The distance between the person and the shuttle is changing at a rate of 640 feet/sec when the shuttle is 4000 ft high and is rising at a rate of 800 ft/sec.

Considering x to be the height of the triangle and y to be the hypotenuse of the triangle, a right-angled triangle will be formed according to the given information having a base of 3000 feet which is the distance of the person from the launch pad.

In this right-angle triangle x will be the height of the shuttle and y will be the distance between the person and the shuttle.

Now we apply the Pythagorean theorem to the triangle;

y² = x²+(3000)²

Now differentiating this equation with respect to time,t :

2y (dy/dt) = 2x(dx/dt) + 0

dy/dt = x(dx/dt)/y

As the shuttle is rising at a speed of 800 ft/sec, (dx/dt)=800

Substitute 4000 for x into the equation, y²=x²+(3000)², to find y;

y² = (4000)²+(3000)²

y = 5000

Now we substitute 4000 for x, 5000 for y, and 800 for dx/dt in the equation, dy/dt = x(dx/dt)/y

dy/dt = 4000(800) / 5000

dy/dt = 640

Therefore, the distance between the person and the shuttle is increasing at a rate of 640 feet/sec.

To learn more about the Pythagorean theorem, click here:

https://brainly.com/question/15819138

#SPJ4

Other Questions