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C38 · Unit C9 Parametric, polar and vector calculus

Parametric derivatives and second derivatives

BC extensionReview editionNot yet human-reviewed
Time
About 45 minutes, untimed
Course framework reference
AP 9.1–9.2 · BC
Builds on
Part of

Progress on this lesson counts in every pathway listed.

Understand the idea

For x equals x(t),y equals y(t), dy over dx equals (dy over dt)/(dx over dt) where dx over dt is not equal to 0. The second derivative is [d over dt(dy over dx)]/(dx over dt), not y seconds (t)/x seconds (t). Horizontal tangents require y minutes (t) equals 0 with x minutes (t) is not equal to 0; vertical tangents require x minutes (t) equals 0 with y minutes (t) is not equal to 0. If both vanish, inspect the local curve separately.

Worked example

For x equals t squared,y equals t cubed and t is not equal to 0, dy over dx equals 3t over 2. Then d squared y over dx squared equals (3 over 2)/(2t) equals 3/(4t). At t equals 2 the point is (4,8), slope 3 and second derivative 3 over 8.

Explore the mathematics

Parametric derivatives and second derivatives

Link two coordinate rates through their common parameter.

Setting 3 of 5

Scroll sideways to see the full diagram.

Both coordinates depend on the same parameter. How do you find dy/dx?xy1x=t², y=t³; t>0

Predict before continuing

Both coordinates depend on the same parameter. How do you find dy/dx?

Read the complete explanation
  1. Both coordinates depend on the same parameter. How do you find dy/dx?
  2. Divide dy/dt=3t² by dx/dt=2t where the denominator is nonzero.
  3. dy/dx=3t/2. Differentiate with respect to t, then divide by dx/dt again for the second derivative.
  4. At t=4, slope=6; second derivative=3/(4t)=0.1875.
See the idea: Parametric derivatives and second derivatives
See the idea: Parametric derivatives and second derivativesFor the t≥0 branch x=t²,y=t³, the point t=2 is (4,8). Its tangent slope is 3 and its second derivative is 3/8. Parameter values t<0 trace a different branch below the x-axis.xy0012.52537.5410t=2

For the t≥0 branch x=t²,y=t³, the point t=2 is (4,8). Its tangent slope is 3 and its second derivative is 3/8. Parameter values t<0 trace a different branch below the x-axis.

Discuss: explain what each label represents, then draw a related example with different values.

A common misconception

The parameter derivative measures change per t; the graph derivative measures change per x. Dividing by x minutes converts between them.

Investigate and explain

Eliminate t in x equals t squared +1,y equals 2t and differentiate implicitly to verify the parametric result.
Identify one step in the worked example, name the mathematical fact that permits it, and state its domain conditions.