Question
Find dy dx for the parametric equations x equals one third t cubed and y =
-t^2 + t. (10 points)
A) -2t + 1/ t^2
B) negative 1 over t squared
C) negative 1 over t cubed
D) 2
-t^2 + t. (10 points)
A) -2t + 1/ t^2
B) negative 1 over t squared
C) negative 1 over t cubed
D) 2
Answers
mathhelper
x = (1/3)t^3 and y = -t^2 + t
dx/dt = t^2 and dy/dt = -2t + 1
dy/dx = (dy/dt)/(dx/dt) = (-2t + 1)/t^2
= -2t/t^2 + 1/t^2
= -2/t + 1/t^2
Unless you have a typo in A), none of the choices are correct.
dx/dt = t^2 and dy/dt = -2t + 1
dy/dx = (dy/dt)/(dx/dt) = (-2t + 1)/t^2
= -2t/t^2 + 1/t^2
= -2/t + 1/t^2
Unless you have a typo in A), none of the choices are correct.