Video: Applications of the Mean Value Theorem

A rock is dropped from a height of 81 ft. Its position 𝑑 seconds after it is dropped until it hits the ground is given by the function 𝑠(𝑑) = βˆ’16𝑑² + 81. Determine how long it will take for the rock to hit the ground. Find the average velocity, 𝑣_(avg), of the rock from the point of release until it hits the ground. Find the time 𝑑 according to the mean value theorem when the instantaneous velocity of the rock is 𝑣_(avg).

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Video Transcript

A rock is dropped from a height of 81 feet. Its position 𝑑 seconds after it is dropped until it hits the ground is given by the function 𝑠 of 𝑑 equals negative 16𝑑 squared plus 81. Determine how long it will take for the rock to hit the ground. Find the average velocity of the rock from the point of release until it hits the ground. And find the time 𝑑 according to the mean value theorem when the instantaneous velocity of the rock is equal to the average velocity.

The rock will reach the ground when its position 𝑠 of 𝑑 is equal to zero. We can therefore set this expression negative 16𝑑 squared plus 81 equal to zero and solve for 𝑑. We add 16𝑑 squared to both sides and then divide through by 16. And we obtain 𝑑 squared to be equal to 81 over 16. We then take the square root of both sides, remembering to take by the positive and negative square root of 81 over 16. And we see that 𝑑 is equal to plus or minus nine-quarters. Now, we can actually disregard negative nine-quarters since we’re working in time. And we find that the rock hits the ground after nine-quarters of a second.

Our next job is to find the average velocity of the rock over this period of time. The definition for average velocity is total displacement divided by time taken. The displacement of our rock is its change in position. That’s negative 81 feet. And it takes nine-quarters of a second to travel this far. So the velocity is negative 81 divided by nine over four. Remember, to divide by a fraction, we can multiply by the reciprocal of that fraction. So we have negative 81 times four over nine. And then we cancel this factor of nine. And so we obtain that the average velocity of our rock is negative 36 feet per second.

For the final part of this question, we’ll need to quote the mean value theorem. Remember, this says that if 𝑓 is a continuous function over some closed interval π‘Ž to 𝑏 and differentiable at every point of that open interval π‘Ž to 𝑏. Then there’s a point 𝑐 in this interval, such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž. We know that the average velocity is negative 36 feet per second. That’s the equivalent of this quotient. The instantaneous velocity can be found by differentiating our function for position. That’s 𝑠 prime of 𝑑 equals negative 32𝑑. In this case then, we can say that 𝑠 prime of 𝑐 is equal to negative 32𝑐. And we obtain the equation negative 32𝑐 equals negative 36. We solve for 𝑐 by dividing both sides by negative 32. And we find that the time at which the instantaneous velocity of the rock is equal to the average velocity is equal to nine-eighths of a second.

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