The Limits of Average Rate of Change: A Calculus Conundrum
A car's position along a straight road is given by the function s(t) = t^3 - 6t^2 + 9t + 5, where s is in meters and t is in seconds, for 0 ≤ t ≤ 5. A student computes the average rate of change of s over the interval [0,5] and gets (s(5)-s(0))/5 = (125-150+45+5 - 5)/5 = (25-5)/5 = 20/5 = 4 m/s. The student then claims that at some instant the car must be traveling at exactly 4 m/s. Another student argues that the car might never actually reach 4 m/s because the average rate of change is just an average. The teacher asks the class to investigate using the Mean Value Theorem and to also consider the instantaneous rate of change at t=2.
1. Verify the student's calculation of the average rate of change over [0,5]. Then determine whether the Mean Value Theorem applies to s(t) on [0,5] and if so, find all values of c in (0,5) such that s'(c) equals the average rate of change.
Read the worked answer
s(0)=5, s(5)=125-150+45+5=25. Average rate = (25-5)/(5-0)=20/5=4 m/s. s(t) is a polynomial, so it is continuous and differentiable everywhere, hence on [0,5] and (0,5). MVT applies. s'(t)=3t^2-12t+9. Set s'(c)=4: 3c^2-12c+9=4 ⇒ 3c^2-12c+5=0. Solve: c = [12 ± sqrt(144-60)]/6 = [12 ± sqrt(84)]/6 = [12 ± 2√21]/6 = 2 ± (√21)/3. Numerically, √21≈4.5826, so c≈2±1.5275, giving c≈3.5275 and c≈0.4725. Both are in (0,5). So there are two times when the instantaneous velocity equals 4 m/s.
2. Find the instantaneous velocity at t=2. Is it equal to the average rate of change over [0,5]? Explain why or why not, and discuss the significance of the MVT result in this context.
Read the worked answer
s'(2)=3(4)-12(2)+9=12-24+9=-3 m/s. This is not equal to 4 m/s. The MVT guarantees at least one point where instantaneous velocity equals average velocity, but not necessarily at t=2. The MVT result shows that the car must have had velocity 4 m/s at some time(s) between 0 and 5, specifically at approximately t=0.47 s and t=3.53 s. At t=2, the car is moving backward at 3 m/s, which is possible because the average velocity is positive but the car can change direction.
3. Suppose the position function were changed to s(t) = t^2 on [0,2]. Would the MVT still apply? Find the value of c guaranteed by the MVT. How does this compare to the previous function?
Read the worked answer
s(t)=t^2 is continuous and differentiable on [0,2]. Average rate = (4-0)/2=2. s'(t)=2t. Set 2c=2 ⇒ c=1. So at t=1, instantaneous velocity equals average velocity. This is simpler because the function is quadratic and the MVT point is the midpoint. In the previous cubic, there were two points because the derivative is quadratic and the horizontal line y=4 intersected it twice. The MVT guarantees at least one, but there can be more.