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1-2 (c) Second derivative of parametric curves

1-2 (c) Second derivative of parametric curves

Point 7 of 8 — Implicit Differentiation, Parametric Differentiation and Higher DerivativesQuestion bank: 0 questions (0 with pictures)

07 1-2 (c) Second derivative of parametric curves

The rule

  • If x=f(t) and y=g(t) then d2ydx2=ddt(dydx)dxdt
  • First find dydx as a function of t, differentiate it with respect to t, then divide by dxdt.
  • A frequent mistake is to write d2y/dt2d2x/dt2; this is wrong.
✍️ Worked examples

Example 18: If x=2n3−5 and y=6n2+1 find d2ydx2 at n=1.

Solution:

  • dxdn=6n2 and dydn=12n.
  • dydx=12n6n2=2n.
  • ddn(2n)=−2n2.
  • d2ydx2=−2/n26n2=−13n4.
  • At n=1 the value is −13.

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