An object moves along a parabolic path defined by the equation y = 2x 2 + 1 in such a way that the y-value is decreasing at a constant rate of 2 units per second. At what rate is the x-coordinate changing when x = 3/2?
I will assume you meant to type
y = 2x^2 + 1
dy/dt = 4x dx/dt
given dy/dt = -2 when x = 3/2
-2 = 4(3/2) dx/dt
-2 = 6 dx/dt
dx/dt = -2/6 = -1/3
post it.
To find the rate at which the x-coordinate is changing, we need to calculate dx/dt, where x is the independent variable and t represents time.
Given that the y-value is decreasing at a constant rate of 2 units per second, we have dy/dt = -2.
We also know that y = 2x^2 + 1. To relate x and y in terms of t, we need to differentiate both sides with respect to t:
dy/dt = d(2x^2 + 1)/dt
Using the chain rule, we have:
-2 = (d(2x^2 + 1)/dx) * (dx/dt)
Simplifying this equation, we get:
(d(2x^2 + 1)/dx) = -2 / (dx/dt)
Now, let's differentiate 2x^2 + 1 with respect to x:
d(2x^2 + 1)/dx = 4x
Substituting this result back into the equation, we have:
4x = -2 / (dx/dt)
Now, we need to find dx/dt when x = 3/2.
Substituting x = 3/2 into the equation, we get:
4 * (3/2) = -2 / (dx/dt)
Simplifying this equation, we find:
6 = -2 / (dx/dt)
Now, we can solve for dx/dt:
(dx/dt) = -2 / 6
Simplifying further, we get:
(dx/dt) = -1/3
Therefore, when x = 3/2, the x-coordinate is changing at a rate of -1/3 units per second.