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Average Rate of Change and Linear Approximation

Created by @nathanedwards
 at November 4th 2023, 9:04:10 pm.
AP_Calculus_AB-Questions
#linear-approximation
#position
#average-rate-of-change

Question

A car is traveling along a straight road and its position, in meters, is given by the function s(t)=3t2−2t+1s(t) = 3t^2 - 2t + 1, where tt is the time in seconds.

a) Find the average rate of change of the car's position with respect to time over the interval [1,2][1, 2].

b) Use a linear approximation to estimate the car's position at t=1.5t = 1.5.

Answer

a) To find the average rate of change of the car's position with respect to time over the interval [1,2][1, 2], we need to calculate the difference in position divided by the difference in time.

The average rate of change is given by:

Average Rate of Change=s(2)−s(1)2−1\text{Average Rate of Change} = \frac{s(2) - s(1)}{2 - 1}

First, we find the position of the car at t=2t = 2:

s(2)=3(2)2−2(2)+1=11s(2) = 3(2)^2 - 2(2) + 1 = 11

Next, we find the position of the car at t=1t = 1:

s(1)=3(1)2−2(1)+1=2s(1) = 3(1)^2 - 2(1) + 1 = 2

Substituting these values into the average rate of change formula:

Average Rate of Change=11−22−1=9\text{Average Rate of Change} = \frac{11 - 2}{2 - 1} = 9

Therefore, the average rate of change of the car's position over the interval [1,2][1, 2] is 9 meters per second.

b) To estimate the car's position at t=1.5t = 1.5 using linear approximation, we can use the tangent line at t=1t = 1. The tangent line at t=1t = 1 will have the same slope as the function s(t)s(t) at that point.

First, let's find the slope at t=1t = 1. We take the derivative of s(t)s(t):

s′(t)=dsdt=ddt(3t2−2t+1)=6t−2s'(t) = \frac{ds}{dt} = \frac{d}{dt}(3t^2 - 2t + 1) = 6t - 2

Evaluating s′(t)s'(t) at t=1t = 1 gives:

s′(1)=6(1)−2=4s'(1) = 6(1) - 2 = 4

Therefore, the slope of the tangent line at t=1t = 1 is 4.

Now, we can form the equation of the tangent line using the point-slope form:

y−y1=m(x−x1)y - y_1 = m(x - x_1)

where (x1,y1)(x_1, y_1) is a point on the line (in this case, (1,s(1))=(1,2)(1, s(1)) = (1, 2)) and mm is the slope. Substituting the values:

y−2=4(x−1)y - 2 = 4(x - 1)

We need to find the value of yy when x=1.5x = 1.5, which corresponds to t=1.5t = 1.5. We substitute x=1.5x = 1.5 into the equation:

y−2=4(1.5−1)y - 2 = 4(1.5 - 1)

Simplifying:

y−2=4(0.5)y - 2 = 4(0.5)
y−2=2y - 2 = 2
y=4y = 4

Therefore, the estimated position of the car at t=1.5t = 1.5 using linear approximation is 4 meters.

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