Calc question...?

Question

A point moves along a curve y=(2x^2)+1 in such a way that the y value is decreasin at the rate of 2 units per second. At what rate is x changing when x= 3/2?

Answer 1

take derivative both sides with respect to t

dy/dt = 4x dx/dt

we want to solve dx/dt

dx/dt = dy/dt /[ 4x]

dy/dt = -2, x = 3/2

so dx/dt = -2/[4*3/2] = -2 /6 = -1/3

rate of x will be decreasing by 1/3 units per second

Answer 2

Well using a little implicit differentiation, we can solve this problem in a cinch.

take the derivative:

dy/dt = 4x(dx/dt) + 0
we know that dy/dt = -2 (y is decreasing by 2 units per second)

so:

-2 = 4(3/2)(dx/dt) solve for dx/dt
-2 = 6(dx/dt)
dx/dt = -2/6 = -1/3

Answer 3

dx/dt = dx/dy * dy/dt. We know dy/dt = -2, so let us find dx/dy:

y = 2x² + 1
1 = 4x dx/dy
dx/dy = 1/(4x)

Since x = 3/2, we have dx/dy = 1/6. So then:

dx/dt = 1/6 * -2 = -1/3

And we are done.