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 tdy/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
收錄日期: 2021-04-12 12:51:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200921083804AAlWsXt