Cogito
AP Calculus BC · Chapter 1 · Lesson 1
Parametric Equations
Two functions of a third variable.
11 problems · about 25 minutes · AP Calculus BC 9.1, 9.2
What this lesson teaches
The student converts between parametric and Cartesian forms and finds dy/dx parametrically.
- A parametric curve gives x and y each as a function of a parameter.
- Eliminating the parameter recovers a Cartesian relation but loses direction.
- dy/dx is the ratio (dy/dt) / (dx/dt).
Warm Up
Straightforward practice. Get the method working first.
4 problemsdx/dt = 3, dy/dt = 15. What is dy/dx?
Answer 5
Why 5.
x = 3t, y = t. What is x when t = 4?
Answer 12
Why 3 × 4.
dx/dt = 4, dy/dt = 12. What is dy/dx?
Answer 3
Why A ratio.
dx/dt = 5, dy/dt = 0. What is dy/dx?
Answer 0
Why Moving horizontally.
Build It Up
The same ideas with more to keep track of.
3 problemsx = t, y = t². Eliminating t gives y = x to what power?
Answer 2
Why Substitute directly.
Select every true statement about parametric curves.
Answer They may cross themselves.; Eliminating the parameter loses the direction.
Why A curve is not a function of x here.
x = t + 2, y = 3t. What is y when x = 5?
Answer 9
Why t = 3.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
4 problemsdx/dt = 0 at a point. Is the tangent vertical there? 1 for yes, 0 for no.
Answer 1
Why The denominator vanishes.
Order the steps for eliminating the parameter.
Answer 1. Solve one equation for t 2. Substitute into the other 3. Simplify to a relation in x and y
Why Solve before substituting.
At a Moment: x = 2t, y = t². What is x when t = 3?
Answer 6
Why 6.
The Slope: dx/dt = 2 and dy/dt = 6. What is dy/dx?
Answer 3
Why 3.