A stone is thrown from the top of a bridge into a lake below such that its height

Question

A stone is thrown from the top of a bridge into a lake below such that its height, h, in

metres, from the lake’s surface, can be modelled by the function h(t) = –4.9t^2 – 8t + 60

a) What is the average rate of change in height of the stone from 1 to 3 seconds after it was

dropped.

b) What is the instantaneous rate of change in height of the stone at 3 seconds? (Carry at

least 3 decimal places.)

c) Use the result from part b to determine the equation of the tangent line to the function

h(t) at 3 seconds? (Recall: y = mx + b)

Answer 1

-4.9t^2 - 8t + 60

h' = -9.8t -8

h'(3) = -9.8(3) - 8 = -29.4 - 8 = - 37.4 meters per second = rate of change after 3 seconds

h'(1) = -9.8 -8 = -17.8 meters per second = rate of change after 1 second

h'(2) = -9.8(2) = -19.6 -8 = -27.6 meters per second = rate of change after 2 seconds

average rate of change from 1 second to 3 seconds = (-37.4-17.8)/2 = -55.2/2 = -27.6 meters per second which also = h'(2) the instantaneous rate of change after 2 seconds. 2 is (3+1)/2 = 4/2 = 2

at 3 seconds the height = -4.9(3)^2 -8(3) + 60 = -44.1-24 + 60 = -20.1 + 60 = 39.9 meters high

that's the point (3, 39.9)

tangent line through that point with slope =h'(3) = -37.4 is

y - 39.9 = -37.4(x-3)

y = -37.4x + 112.2 +39.9

y = -37.4x + 152.1