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M14056009 Key Questions

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1. 7.5/8
The height in metres of a ball dropped from the top of the CN Tower is given by h(t)= -4.9t2+450, where t is time elapsed in seconds.
(a) Draw the graph of h with respect to time

(b) Find the average velocity for the first 2 seconds after the ball was dropped h(0)=(0,450), h(2)=(2,430.4)
= (430.4-450)/(2-0)
= -9.8m/s √
(c) Find the average velocity for the following time intervals
(1) 1 ≤ t ≤ 4 h(1)=(1,445.1) h(4)=(4,371.6)
= (371.6-445.1)/(4-1)
= -24.5m/s √
(2) 1 ≤ t ≤ 2 h(1)=(1,445.1) h(2)=(2,430.4)
= (430.4-445.1)/(2-1)
= -14.7m/s √

(3) 1 ≤ t ≤ 1.5 h(1)=(1,445.1) h(1.5)=(1.5, 438.98)
= (438.98-445.1)/(1.5-1)
= -12.25m/s √
(d) Use the secant method to approximate the instantaneous velocity at t=1 h(0.5) = (0.5, 448.78) …show more content…

H(2)= -5(2)2+20(2)+1 = -20+40+1 =21 H(2+h)= -5(2+h)2 + 20(2+h) +1 = -5(4+4h+h2) + 40+20h+1 = -20-20h-5h2+40+20h+1= -5h2+21 lim(h->0) H(2+h)-H(2)/h = -5h2+21-21/h = -5h = -5(0) = 0 √
(b) A particle’s motion is described by the equation d=t2-8t+15 where d and t are measured in metres and seconds. Show that the particle is at rest when t = 4.
D(4)= (4)2-8(4)+15 =16-32+15= -1 D(4+h)= (4+h)2-8(4+h)+15= 16+8h+h2-32-8h+15= h2-1 limD(4+h)-D(4)/h = h2-1+1/h = h =0 √
6. 4/4
For the following graph

(a) Determine the intervals between which the rate of change is positive and negative. The function is increasing at x<-1 and x>1, hence its rate of change is positive. The function is decreasing at -1<x<1, hence its rate of change is negative. √
(b) State where the rate of change is zero. The instantaneous rate of change is zero at x=-1 and x=1 √
(c) List the local maximums and minimums of the function. F(-1) is a local maximum and F(1) is a local minimum. √
7. 10/10
For the function f(x) =2x3-7x2+4x+1 (a) Find the instantaneous rate of change at x=0 and x=1 (1) F(0+h) = 2(h)3 – 7(h)2+4(h)+1 =2h3-7h2+4h+1 F(0) = 1 lim(h->0) (F(0+h)-F(0))/h = (2h3-7h2+4h+1-1)/h = (2h3-7h2+4h)/h = 2h2-7h+4 = 4 ∴the instantaneous rate of change at x=0 is 4 √

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