1. What is the point that corresponds to t = 90 when x = 4 sin t and y = 3 cos t?
2. The path of a shotput thrown is represented by the following parametric equations. Distance is
measured in feet and time in seconds. Rewrite the parametric equations by eliminating the
parameter.
x(t) = 23t
y(t) = 5 + 19.3t − 16t^2
3. Write the equation for the curve described by the parametric equations x = 5 − cos t and y =
2 + 2 sin t in rectangular form.
4. x(t) = t^2
y(t) = 2t + 1
Rewrite y(t) when x(t) = t and again when x(t) = 2t
1. When t = 90 degrees (or pi/2 radians), x = 4 sin(90) = 4 * 1 = 4 and y = 3 cos(90) = 3 * 0 = 0. So the point is (4, 0).
2. We can solve for t in the first equation: t = x/23. Then substitute that into the second equation: y = 5 + 19.3(x/23) - 16(x/23)^2. Simplifying, we get y = (747x - 736x^2)/529.
3. We can use the identity cos^2 t + sin^2 t = 1 to eliminate cos t from the first equation: cos t = 5 - x. Then substitute that into the second equation: y = 2 + 2 sin t = 2 + 2 sqrt(1 - cos^2 t) = 2 + 2 sqrt(1 - (5 - x)^2). Simplifying, we get y = 2 + 2 sqrt(4x - x^2 - 21).
4. When x(t) = t, we can solve for t: t = sqrt(x). Then substitute that into the second equation: y = 2(sqrt(x)) + 1 = 2x^(1/2) + 1. When x(t) = 2t, we can solve for t: t = x/2. Then substitute that into the second equation: y = 2(x/2) + 1 = x + 1.