Math: Parametric Equations and Polar Coordinates Worksheet

Question 1

A curve is defined by x = 3t + 1 and y = 2t - 4. Eliminate the parameter t and write the Cartesian equation of the curve.

Accepted Answer

From x = 3t + 1, we get t = (x - 1)/3. Substitute into y = 2t - 4 to get y = 2(x - 1)/3 - 4, which simplifies to y = (2/3)x - 14/3.

Question 2

A particle moves according to x = 4 cos t and y = 4 sin t. Write a Cartesian equation for its path and describe the graph.

Accepted Answer

Using x = 4 cos t and y = 4 sin t, we have x^2 = 16 cos^2 t and y^2 = 16 sin^2 t. Adding gives x^2 + y^2 = 16. The graph is a circle centered at the origin with radius 4.

Question 3

For the parametric equations x = t^2 + 1 and y = t - 3, eliminate the parameter t and write the equation in terms of x and y.

Accepted Answer

From y = t - 3, we get t = y + 3. Substitute into x = t^2 + 1 to get x = (y + 3)^2 + 1.

Question 4

A point has polar coordinates (5, pi/6). Convert the point to Cartesian coordinates.

Accepted Answer

Use x = r cos theta and y = r sin theta. Then x = 5 cos(pi/6) = 5 root 3 / 2 and y = 5 sin(pi/6) = 5/2. The Cartesian coordinates are (5 root 3 / 2, 5/2).

Question 5

Convert the Cartesian point (-3, 3 root 3) to polar coordinates with r > 0 and 0 less than or equal to theta less than 2pi.

Accepted Answer

First find r = root((-3)^2 + (3 root 3)^2) = root(9 + 27) = 6. Since x is negative and y is positive, the point is in Quadrant II. The reference angle is pi/3, so theta = 2pi/3. The polar coordinates are (6, 2pi/3).

Question 6

Rewrite the polar equation r = 6 cos theta in Cartesian form and identify the graph.

Accepted Answer

Multiply both sides by r to get r^2 = 6r cos theta. Since r^2 = x^2 + y^2 and r cos theta = x, the equation becomes x^2 + y^2 = 6x. Completing the square gives (x - 3)^2 + y^2 = 9. The graph is a circle centered at (3, 0) with radius 3.

Question 7

Rewrite the polar equation r = 4 sin theta in Cartesian form and identify the graph.

Accepted Answer

Multiply both sides by r to get r^2 = 4r sin theta. Replace r^2 with x^2 + y^2 and r sin theta with y to get x^2 + y^2 = 4y. Completing the square gives x^2 + (y - 2)^2 = 4. The graph is a circle centered at (0, 2) with radius 2.

Question 8

Find a polar equation for the circle x^2 + y^2 = 25.

Accepted Answer

Since x^2 + y^2 = r^2, the equation becomes r^2 = 25. With positive radius, this is r = 5. This represents a circle centered at the origin with radius 5.

Question 9

Find a polar equation for the line y = x.

Accepted Answer

Use x = r cos theta and y = r sin theta. Then r sin theta = r cos theta. For points not at the origin, divide by r to get sin theta = cos theta, so tan theta = 1. Therefore theta = pi/4 is a polar equation for the line.

Question 10

A particle moves with x = 2 cos t and y = 3 sin t for 0 less than or equal to t less than 2pi. Write a Cartesian equation for the path and describe the graph.

Accepted Answer

From x = 2 cos t, we get x^2/4 = cos^2 t. From y = 3 sin t, we get y^2/9 = sin^2 t. Adding gives x^2/4 + y^2/9 = 1. The graph is an ellipse centered at the origin with horizontal semi-axis 2 and vertical semi-axis 3.

Question 11

For the polar equation r = 2 + 2 cos theta, find the value of r when theta = 0, pi/2, and pi.

Accepted Answer

When theta = 0, r = 2 + 2 cos 0 = 4. When theta = pi/2, r = 2 + 2 cos(pi/2) = 2. When theta = pi, r = 2 + 2 cos pi = 0. So the values are 4, 2, and 0.

Question 12

A point is given in polar form as (-4, pi/3). Rewrite the point using a positive radius.

Accepted Answer

A negative radius means move in the opposite direction, so add pi to the angle. The equivalent point is (4, 4pi/3).

Question 13

Determine the slope of the line tangent to the parametric curve x = t^2 + 1 and y = 3t - 2 at t = 2.

Accepted Answer

For a parametric curve, dy/dx = (dy/dt) / (dx/dt). Here dx/dt = 2t and dy/dt = 3. At t = 2, dy/dx = 3/4. The slope of the tangent line is 3/4.

Question 14

Find dy/dx for the parametric equations x = sin t and y = cos t.

Accepted Answer

Differentiate to get dx/dt = cos t and dy/dt = -sin t. Then dy/dx = (-sin t) / (cos t) = -tan t, as long as cos t is not zero.

Question 15

Convert the Cartesian equation x = -2 to a polar equation.

Accepted Answer

Use x = r cos theta. Then r cos theta = -2. This is a polar equation for the line x = -2.

Math: Parametric Equations and Polar Coordinates

Math: Parametric Equations and Polar Coordinates

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