Asked by Alice
Find d^2y/dx^2 for the curve given by x = 2t + 5 and y= 3t/2
a) 3/(4t+10)
b) (4t+10)/3
c) 1
d) 0
a) 3/(4t+10)
b) (4t+10)/3
c) 1
d) 0
Answers
Answered by
Reiny
x = 2t + 5
t = (x-5)/2
and y= 3t/2 = (3/2)(x-5)/2 = (3/4)(x - 5)
or
y = (3/4)x - 15/4
so your given parametric equations are simply the straight line
y = (3/4)x - 15/4
dy/dx = 3/4
d(dy/dx) / dx = 0
t = (x-5)/2
and y= 3t/2 = (3/2)(x-5)/2 = (3/4)(x - 5)
or
y = (3/4)x - 15/4
so your given parametric equations are simply the straight line
y = (3/4)x - 15/4
dy/dx = 3/4
d(dy/dx) / dx = 0
Answered by
oobleck
Since x and y are both linear functions of t,
dy/dx = (dy/dt) / (dx/dt) will be a constant.
d^2y/dx^2 = 0
dy/dx = (dy/dt) / (dx/dt) will be a constant.
d^2y/dx^2 = 0