Let f be the function given by f(x)=3sqrt(x-2).
A) On the axes provided sketch graph of f and shade the region R enclosed by the graph f, the x-axis, and the vertical line x=8 (already completed just mentioned it for part B.)
B) Find area of the region R described in part (A).
C) Rather than using line x=8 as in part (A),consider the line x=k, where k can be any number greater than 2. Let A(k) be the area of the region enclosed by the graphs of f, the x-axis, and the vertical line x=k. Write and integral expression for A(k).
D) Let A(k) be as described in part (C). Find the rate of change of A with respect to k when k=8.
A) To sketch the graph of the function f(x) = 3√(x-2), we start by finding some key points.
- Substitute x = 2 to find the y-intercept: f(2) = 3√(2-2) = 0, so the point (2, 0) is on the graph.
- Find some additional points by substituting different values of x. For example, if we substitute x = 5, we get f(5) = 3√(5-2) = 3√3 ≈ 5.196. So, the point (5, 5.196) is on the graph.
Now, plot these points on the coordinate axes and try to connect them smoothly. Since the function involves a square root, it is always non-negative, meaning the graph lies above or on the x-axis between x = 2 and x = 8. Shade the region R enclosed by the graph, the x-axis, and the vertical line x = 8 as directed.
B) To find the area of the region R, we can divide it into two parts: a rectangle and a triangle.
- The rectangle has a base of 8 - 2 = 6 (the difference between the x-values) and a height of f(2) = 0 (the y-value at x = 2). So, the area of the rectangle is 6 * 0 = 0.
- The triangle has a base of 8 - 2 = 6 (the difference between the x-values) and a height of f(8) = 3√(8-2) = 3√6 ≈ 7.348. So, the area of the triangle is (1/2) * 6 * 7.348 ≈ 22.044.
Therefore, the total area of the region R is 0 + 22.044 = 22.044 square units.
C) To write an integral expression for the area A(k), we consider the region enclosed by the graph of f, the x-axis, and the vertical line x = k (where k > 2).
We can find the area by integrating the function f(x) from x = 2 to x = k:
A(k) = ∫[2 to k] 3√(x-2) dx
D) To find the rate of change of A with respect to k when k = 8, we differentiate the integral expression A(k) with respect to k.
dA/dk = d/dk ∫[2 to k] 3√(x-2) dx
Applying the Fundamental Theorem of Calculus, we get:
dA/dk = 3√(k-2)
Now, substitute k = 8 into the expression to find the rate of change:
dA/dk = 3√(8-2) = 3√6 ≈ 7.348
Therefore, when k = 8, the rate of change of A with respect to k is approximately 7.348 square units per unit change in k.