Problem 1 Suppose we have a parametric curve described by x(t) = e^-t, y(t) = +

Question

Problem 1 Suppose we have a parametric curve described by x(t) = e^-t, y(t) = + 2, with t greater than equal to 0. What does the graph of this look like? Remember to include the domain of the parameter in your analysis! A) The line with a slope of 1 and y-intercept 2 B) The line with a slope of -1 and y-intercept 2 C) The ray with a slope of 1 and y-intercept 2 starting at (1,3) and extending in the first quadrant D) The ray with a slope of 1 and y-intercept 2 starting at (0,2) and extending in the first quadrant E) The line segment between (0,2) and (1,3) F) None of these. Problem 2 The slope of the tangent line to a polar curve is NOT r'(theta). To get the slope of the tangent line, you must first convert from polar to parametric (or rectangular, but this is harder). Then you differentiate and so on. Find the slope of the line tangent to the graph of r = 4 cos(2theta) at the point/angle where theta = Pi/2. A) 0 B) 4 C) -4 D) -8 E) -2 F) Vertical tangent (i.e., slope is undefined) Problem 3 What is the length of the graph of from x = 0 to x = 2? f (x) = -1/6 (x^2 + 4)^3/2 A)10 /3 B 15/2 C 44 / 3 D3/44 E) 8/3

Explanation / Answer

Problem 1

y = x + 2

So it is aline with slope 1 and y intercept 2. (A is the correct answer)

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