AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/9/9-1)
Unit 9 · Topic 9.1
BC only9.1 Defining and Differentiating Parametric Equations
BC only. Parametric equations describe a curve by giving x and y separately as functions of a third variable, usually t. The slope of the curve is dy/dx = (dy/dt)/(dx/dt), as long as dx/dt isn't zero.
Key terms
- parameter
- parametric equations
- dy/dx
- horizontal tangent
- vertical tangent
What parametric equations are
This whole unit is BC only. Instead of writing y as a function of x, you give both coordinates as functions of a parameter t: x = x(t) and y = y(t). As t changes, the point (x(t), y(t)) traces a curve. Think of t as time and the point as a bug crawling across the plane.
Parametric curves can do things ordinary functions can't: loop back, cross themselves, or run straight up. For example, x = cos t, y = sin t for 0 ≤ t ≤ 2π traces the unit circle once, counterclockwise, starting at (1, 0).
Sometimes you can eliminate the parameter to get an equation in x and y. If x = 2 cos t and y = 3 sin t, then (x/2)² + (y/3)² = cos² t + sin² t = 1, which is an ellipse. You won't always need to do this, but it helps you picture the curve.
Slope of a parametric curve
By the chain rule, dy/dt = (dy/dx)(dx/dt). Solve for dy/dx:
dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0.
This is the slope of the tangent line at the point the curve reaches at time t. To write the tangent line, find the point (x(t), y(t)) and the slope at that same t, then use point-slope form.
Horizontal and vertical tangents
A horizontal tangent needs the curve to stop rising or falling while it still moves sideways, and a vertical tangent is the reverse (see the table). When both derivatives are zero at the same t, the formula gives 0/0 and doesn't decide anything. The curve might have a cusp (a sharp point) there, or a tangent you'd need limits to find. Exam questions rarely go further than recognizing that case.
| Tangent line | Condition |
|---|---|
| horizontal | dy/dt = 0 and dx/dt ≠ 0 |
| vertical | dx/dt = 0 and dy/dt ≠ 0 |
| can't tell from these alone | dy/dt = 0 and dx/dt = 0 |
Direction of motion
The signs of dx/dt and dy/dt tell you which way the point is moving. dx/dt > 0 means moving right, dx/dt < 0 means moving left. dy/dt > 0 means moving up, dy/dt < 0 means moving down. These signs are separate from the slope: a slope of 1 could mean moving up and right, or down and left.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Slope, tangent line, horizontal and vertical tangents
A curve is given by x = t² − 1 and y = t³ − 3t. (a) Find dy/dx. (b) Write the tangent line at t = 2. (c) Find the points where the tangent line is horizontal or vertical.
Show the solutionHide the solution
- Step 1: (a) dx/dt = 2t and dy/dt = 3t² − 3. So dy/dx = (3t² − 3)/(2t), for t ≠ 0.
- Step 2: (b) At t = 2: the point is (4 − 1, 8 − 6) = (3, 2), and the slope is (12 − 3)/4 = 9/4.
- Step 3: Tangent line: y − 2 = (9/4)(x − 3).
- Step 4: (c) Horizontal: dy/dt = 3t² − 3 = 0 at t = ±1, and dx/dt = ±2 ≠ 0 there. Points: t = 1 gives (0, −2); t = −1 gives (0, 2).
- Step 5: Vertical: dx/dt = 2t = 0 at t = 0, and dy/dt = −3 ≠ 0 there. Point: (−1, 0).
Answer: (a) dy/dx = (3t² − 3)/(2t) (b) y − 2 = (9/4)(x − 3) (c) Horizontal tangents at (0, −2) and (0, 2); vertical tangent at (−1, 0).
- Example 2
Trap: dividing in the wrong order
For x = 2 cos t, y = 3 sin t, find dy/dx at t = π/4.
Show the solutionHide the solution
- Step 1: dx/dt = −2 sin t and dy/dt = 3 cos t.
- Step 2: dy/dx = (dy/dt)/(dx/dt) = (3 cos t)/(−2 sin t).
- Step 3: At t = π/4, sin t = cos t = √2/2, so dy/dx = 3/(−2) = −3/2.
- Step 4: Common wrong answer: −2/3, from dividing dx/dt by dy/dt.
Answer: dy/dx = −3/2
Common mistakes
- Computing (dx/dt)/(dy/dt) instead of (dy/dt)/(dx/dt).
- Using a t-value as if it were an x-value when writing the tangent line. Convert t to the point (x(t), y(t)) first.
- Declaring a horizontal tangent where dy/dt = 0 without checking that dx/dt ≠ 0.
- Confusing the direction of motion with the slope.
On the exam
- BC free-response questions often combine parametric derivatives with motion (9.6). Slope at a time, tangent line and horizontal/vertical tangent points are common parts.
- Give points as ordered pairs (x, y), not as t-values, unless the question asks for t.
Connected topics
Videos
Check yourself
4 questions on 9.1 Defining and Differentiating Parametric Equations. Pick an answer to see if you got it, and why.
A curve is defined by the parametric equations x = t² − 1 and y = t³ − 3t. What is the slope of the line tangent to the curve at the point where t = 2?
A curve is given by x = eᵗ and y = t² + 3t. Which of the following is an equation of the line tangent to the curve at the point where t = 0?
A curve is defined by x = t³ − 3t and y = t² − 4t. At which points on the curve is the tangent line vertical?
A curve is defined by x = 2t − sin(3t) and y = t² + cos(2t). For 0 < t < 2, the curve has a horizontal tangent line at exactly one point. What is the x-coordinate of that point?
0 of 4 answered