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Second Derivative for Parametric Curves
Worksheet
Grade Level
Calculus II
CAL2.95 Second Derivative for Parametric Curves
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A path is modeled by x(t) = e
t
- t and y(t) = t
3
+ t, where x and y are measured in meters. At t = 1, which statement correctly gives d²y/dx² and interprets the path's concavity as y changes with x?
A
d²y/dx² = 6·t; at t = 1, it is 6, so the curve is concave up.
B
d²y/dx² = -(-6·t·(exp(t) - 1) + (3·t
2
+ 1)·exp(t))/(exp(t) - 1)
3
; at t = 1, it is 2·(-e + 3)/(-e + 1)
3
, so the curve is concave down.
C
d²y/dx² = (-6·t·(-exp(t) + 1) - (3·t
2
+ 1)·exp(t))/(-exp(t) + 1)
2
; at t = 1, it is 2·(-3 + e)/(-e + 1)
2
, so the curve is concave down.
D
d²y/dx² = -(6·t·(exp(t) - 1) - (3·t
2
+ 1)·exp(t))/(exp(t) - 1)
3
; at t = 1, it is 2·(-3 + e)/(-e + 1)
3
, so the curve is concave up.
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