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Math  /  Calculus

Question4. The path of a baseball relative to the ground can be modelled by the function d(t)=−t2+8t+1d(t)=-t^{2}+8 t+1 where d(t)d(t) represents the height of the ball in metres after tt seconds. a. Find the average rate of change of the ball between 1 and 3 seconds. [4 marks] b. Using the secant method, find the instantaneous rate of change at 2 seconds. [5 marks]

Studdy Solution

STEP 1

1. The function d(t)=−t2+8t+1d(t) = -t^2 + 8t + 1 models the height of the baseball over time.
2. The average rate of change between two points t=at=a and t=bt=b can be found using the formula d(b)−d(a)b−a\frac{d(b) - d(a)}{b - a}.
3. The instantaneous rate of change at t=ct=c can be approximated using the secant method, which involves calculating the average rate of change over increasingly smaller intervals around t=ct=c.

STEP 2

1. Calculate the average rate of change of d(t)d(t) between t=1t=1 and t=3t=3.
2. Set up the secant method to approximate the instantaneous rate of change at t=2t=2.
3. Calculate the instantaneous rate of change at t=2t=2 using the secant method.

STEP 3

Calculate d(1)d(1) using the function d(t)=−t2+8t+1d(t) = -t^2 + 8t + 1.
d(1)=−(1)2+8(1)+1=−1+8+1=8 d(1) = -(1)^2 + 8(1) + 1 = -1 + 8 + 1 = 8

STEP 4

Calculate d(3)d(3) using the function d(t)=−t2+8t+1d(t) = -t^2 + 8t + 1.
d(3)=−(3)2+8(3)+1=−9+24+1=16 d(3) = -(3)^2 + 8(3) + 1 = -9 + 24 + 1 = 16

STEP 5

Calculate the average rate of change between t=1t=1 and t=3t=3 using the formula d(b)−d(a)b−a\frac{d(b) - d(a)}{b - a}.
\text{Average rate of change} = \frac{d(3) - d(1)}{3 - 1} = \frac{16 - 8}{3 - 1} = \frac{8}{2} = 4 $ \text{ meters per second}

STEP 6

Set up the secant method to approximate the instantaneous rate of change at t=2t=2. Choose values close to 2, say t=2t=2 and t=2+ht=2+h where hh is a small increment.
Use h=0.01 for a small increment. \text{Use } h = 0.01 \text{ for a small increment.}

STEP 7

Calculate d(2)d(2) using the function d(t)=−t2+8t+1d(t) = -t^2 + 8t + 1.
d(2)=−(2)2+8(2)+1=−4+16+1=13 d(2) = -(2)^2 + 8(2) + 1 = -4 + 16 + 1 = 13

STEP 8

Calculate d(2+0.01)d(2 + 0.01) using the function d(t)=−t2+8t+1d(t) = -t^2 + 8t + 1.
d(2.01)=−(2.01)2+8(2.01)+1=−4.0401+16.08+1=13.0399 d(2.01) = -(2.01)^2 + 8(2.01) + 1 = -4.0401 + 16.08 + 1 = 13.0399

STEP 9

Calculate the instantaneous rate of change at t=2t=2 using the secant method formula d(2+h)−d(2)h\frac{d(2+h) - d(2)}{h} with h=0.01h = 0.01.
Instantaneous rate of change=d(2.01)−d(2)0.01=13.0399−130.01=0.03990.01=3.99 meters per second \text{Instantaneous rate of change} = \frac{d(2.01) - d(2)}{0.01} = \frac{13.0399 - 13}{0.01} = \frac{0.0399}{0.01} = 3.99 \text{ meters per second}
Solution: a. The average rate of change of the ball between 1 and 3 seconds is 44 meters per second. b. Using the secant method, the instantaneous rate of change at 2 seconds is approximately 3.993.99 meters per second.

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