A point moves along the curve y=2x^2+1

1. Nov 11, 2014

dnt

• Thread was started in a non-homework section, so is missing the template
A point moves along the curve y=2x^2+1 in such a way that the y value is decreasing at the rate of 2 units per second. At what rate is x changing when x=(3/2)?

Ok here's what I've done...

Dy/dt=(4x)(dx/dt)

-2 = (4*3/2)(dx/dt)

So dx/dt=-1/3 (ie decreasing by 1/3 units per second)

First, is that right? Second when I look at the graph and y is decreasing it's the left side of the parabola however as I go down the graph the x increasing from left to right. So either my work is wrong above or I am misinterpreting the graph. What have I done wrong? Thanks.

2. Nov 11, 2014

pasmith

Yes.

Looking at a graph of $y$ against $x$ only tells you how $y$ varies with $x$. That graph tells you nothing about how $y$ varies with $t$ unless you know how $x$ varies with $t$. If $y$ increases with $x$ but $x$ itself is decreasing with $t$, then $y$ will decrease with $t$.

3. Nov 11, 2014

PeroK

In which direction do you think the particle is moving? (You've not done anything wrong.)