# Math Help - calc 1 question

1. ## calc 1 question

point moves along a curve y=2x^2+1 in such a way that the y value is decreasing at the rate of 2 units per sec. at what rate is x changing when x=3/2.

y' = 4x so F'(3/2)= m = 6
so slope=6 and f(3/2) = y= 11/2

point slope formula = y-(11/2) = 6(x-(3/2)) = y= 12x/2 -7/2

so do I set x=0 to find rate of change for x?

2. Originally Posted by weezie23
point moves along a curve y=2x^2+1 in such a way that the y value is decreasing at the rate of 2 units per sec. at what rate is x changing when x=3/2.

y' = 4x so F'(3/2)= m = 6
so slope=6 and f(3/2) = y= 11/2

point slope formula = y-(11/2) = 6(x-(3/2)) = y= 12x/2 -7/2

so do I set x=0 to find rate of change for x?
nope. implicit differentiation is what is needed here. what you are doing is for a whole other class of problem. are you sure you're getting what's going on in class?

$y = 2x^2 + 1$ ..............differentiate with respect to time

$\Rightarrow \frac {dy}{dt} = 4x \frac {dx}{dt}$

now, you are told $\frac {dy}{dt} = -2, \mbox{ and } x = \frac 32$, you are asked for $\frac {dx}{dt}$, just solve for it