Calculus: How do I interpret this shaded region?

Question

Two cars A and B start side by side and accelerate from rest. The function representing the velocity of A looks like the square root of x starting from (0,0) and the function representing the velocity of B looks like the cube root of x starting from (0,0). At one minute, the graphs of the two functions intersect.

a. Which car is ahead after one minute?

b. What is the meaning of the area of the region enclosed by the two curves?

c. Which car is ahead after two minutes?

d. Estimate the time when the cars are again side by side.

Answer 1

Well, for the first minute B is always moving faster than A. So guess which one is ahead at the end of that time?

The integral of the speed is the distance covered. The shaded area indicates the difference between the integrals, and so the distance between the cars.

To do c and d you need to actually integrate their speed over time so as to get the distance traveled by each particular time.

Answer 2

Assuming the velocity functions are functions of time, and that time is measured in minutes:

a) Car B. For numbers in between 0 and 1, the cube root will be bigger than the square root (e.g., ³√(1/64) = 1/4, but √(1/64) = 1/8). So up until t=1, car B is going faster, thus is ahead of car A.

b) When graphing velocity against time, the area under the curve on an interval shows the distance traveled between the two times. So the enclosed area is just the area under the B curve minus the area under the A curve. This would then represent how far B is from A.

c) Take the integral of each function from 0 to 2. Calculate these out to see who has the bigger area, and has thus covered the most distance from the start.

b) Take the general intergrals of both, set them equal to each other, and solve for t. I get about 2.027 seconds

Answer 3

If t is in minutes,
Aa = (1/2)Kt^-(1/2)
Va = Kt^(1/2)
Sa = (2/3)Kt^(3/2)
Ab = (1/3)Kt^-(2/3)
Vb = Kt^(1/3)
Sb = (3/4)Kt^(4/3)
a. At 1 minute B is ahead of A (3/4 > 2/3)
b. The area between the curves represents the distance between the cars.
c. At 2 minutes
Sa = 1.8856K
Sb = 1.8899K
B is very slightly ahead of A
d. When the cars are again side-by-side,
(2/3)Kt^(3/2) = (3/4)Kt^(4/3)
t^(3/2) = (9/8)t^(4/3)
(3/2)Ln(t) = Ln(9/8) + (4/3)Ln(t)
[3/2 - 4/3]Ln(t) = Ln(9/8)
[(9 - 8)/6]Ln(t) = Ln(9/8)
Ln(t) = 6Ln(9/8)
t = (9/8)^6
t ≈ 2.0273 min. ≈ 2 min. 1.62 sec.

Answer 4

apparently you don't realize that the integral of velocity is displacement