Help! I'm looking for the second derivative of a parametric equation?

2020-09-21 4:38 pm
Hi. Can you help please? What is the second derivative of the parametric equation when x = sin t and y = 3 cos t. The answers say -3 sec^3 t but I get something totally different.Please help Yahooland.

回答 (1)

2020-09-21 5:10 pm
✔ 最佳答案
y = 3 cos t
dy/dt = -3 sin t

x = sin t
dx/dt = cos t

dy/dx
= (dy/dt)/(dx/dt)
= - 3 sin t / cos t
= - 3 tan t

d²y/dx²
= (d/dx) (-3 tan t)
= -3 sec² t (dt/dx)
= -3 sec² t / (dx/dt)
= -3 sec² t / cos t
= -3 sec² t / (1/sec t)
= -3 sec² t × sec t
= -3 sec³ t


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