Recall that the graph of x^2+y^2 = 1 is a circle with center (0, 0) and radius 1. Then what would the graph of (x−a)^2+ (y−b)^2 = r^2 be?

Question

2. Suppose that v(t) = sqrt (9 − (t − 3)^2) tells us the instantaneous velocity of a moving object on the interval 0 ≤ t ≤ 6, where t is measured in minutes and v is measured in meters per minute.

(a) Recall that the graph of x^2+y^2 = 1 is a circle with center (0, 0) and radius 1. Then what would the graph of (x−a)^2+ (y−b)^2 = r^2 be? Describe this as accurately as possible. Now, rewrite the equation for v(t) in the form (v(t)−a)^2 + (t−b)^2 = r^2 . Determine a, b, r.

(b) Use Part 2a to sketch an accurate graph of y = v(t). Caution! Do you have a full circle?

(c) Using your picture, evaluate [0,6] v(t)dt exactly.

(d) In terms of the physical problem of the moving object with velocity v(t), what is the meaning of [0,6] v(t)dt? Include units on your answer.

(e) Determine the exact average value of v(t) on [0, 6]. Include units on your answer.

(f) On the same graph that you used in part 2a, sketch a rectangle whose base is the line segment from t = 0 to t = 6 on the t-axis such that the rectangle’s area is equal to the value of [0,6] v(t)dt. What is the rectangle’s exact height?

(g) How can you use the average value you found in Part 2e to compute the total distance traveled by the moving object over [0, 6]?

Answer 1

I will answer one part for you.

(x-a)² + (y-b)² = r²

is the equation of a circle with center at (a,b) and radius r.

The rest of the problem is good exercise for you!