Another Expression There is another way to define slope of a tangent: We let h = x – aWe let h = x – a Then x = a + hThen x = a + h Thus the slope of secant line PQ isThus the slope of secant line PQ is
Taking it to the limit we get... Instantaneous Velocity This means that the velocity at time t = a is equal to thevelocity at time t = a is equal to the slope of the tangent line at P.slope of the tangent line at P.
Example (Calculator active) Suppose a ball is dropped from the top of a tower 450m. Tall. Find the velocity at 5 seconds. Position function f(t) = 4.9t 2
Example (cont’d) a) The velocity after 5 s is v(t)=9.8t v(5) = (9.8)(5) = 49 m/s.
Part deux: How fast is the ball going when it hits the ground?
Example (cont’d) It hits the ground when it traveled 450m The ball will hit the ground when, 4.9t 2 = 450. Solving for t gives t ≈ 9.6 s. The velocity of the ball as it hits the ground is therefore v(t) = 9.8t ≈ 94 m/s
Example Temperature readings T (in °C) were recorded every hour starting at midnight on a day in Whitefish, Montana. The time x is measured in hours from midnight. The data are given in the table on the next slide:
a) Find the average rate of change of temperature with respect to time i.from noon to 3 P.M. ii.from noon to 2 P.M. iii.from noon to 1 P.M. b) Estimate the instantaneous rate of change at noon.
Solution to (a) i. From noon to 3 P.M. the temperature changes from 14.3 °C to 18.2 °C, so ∆T = T(15) – T(12) = 18.2 – 14.3 = 3.9 °C while the change in time is ∆x = 3 h. Therefore, the average rate of change of temperature with respect to time is
Solution to (a) (cont’d) ii. From noon to 2 P.M. the average rate of change is iii. From noon to 1 P.M. the average rate of change is
Solution to (b) We plot the given data on the next slide and sketch a smooth curve of the temperature function. Then we draw the tangent at the point P where x = 12. By measuring the sides of triangle ABC, we estimate the slope of the tangent line to be 10.3/5.5 ≈ 1.9.