Calculus BC: Applications of the Derivative

Problem : Suppose a rock is thrown straight up from atop a 200 -meter-high cliff at an initial speed of 30 feet per second. The height, in meters, of the rock above the ground (until it lands) at time t is given by the function h(t) = - gt ²/2 + 30t + 200 , where g 9.81 is a constant of gravitational acceleration. When does the rock reach its maximum height? What is this maximum height? How fast is the rock moving after 3 seconds?

Solution for Problem 1 >>

When the rock reaches its maximum height, it is instantaneously stationary, with speed 0 . Solving

h'(t) = - gt + 30 = 0

for t , we obtain t = 30/g 3.06 as the time when the rock reaches its maximum height. Substituting back into h(t) , we find that the maximum height is

h(30/g) = +30 +200 = +200 245.89

measured in meters. To find the speed at time t = 3 , we compute

h'(3) = (- g)(3) + 30 0.58

meters per second, which makes sense, because the rock is about 0.06 seconds away from reaching its maximum height and coming to an instantaneous stop.

Close

Problem : The position of a box, in a certain coordinate system, attached to the end of a spring is given by p(t) = sin(2t) . What is the acceleration of the box at time t ? How does this relate to its position?

Problem : Suppose the velocity of a sprinter (in meters per second) at time t seconds after the start of a 40 meter dash is given by

Solution for Problem 3 >>

The acceleration is given by

v'(t) =

at time t . Thus the acceleration at time t = 1 is 3/2 meters per second per second.

Close