# If x^2+y^2=25 and dy/dt=6, how do you find dx/dt when y=4 ?

Sep 21, 2014

There are two values depending on the point.

$\frac{\mathrm{dx}}{\mathrm{dt}} = \left\{\begin{matrix}- 8 \text{ at " (34) \\ 8 " at } \left(- 3 4\right)\end{matrix}\right.$

Let us look at some details.

First, let us find the values of $x$.

By plugging in $y = 4$ into ${x}^{2} + {y}^{2} = 25$,

${x}^{2} + 16 = 25 R i g h t a r r o w {x}^{2} = 9 R i g h t a r r o w x = \pm 3$

Now, let us find some derivatives.

By differentiating with respect to $t$,

$\frac{d}{\mathrm{dt}} \left({x}^{2} + {y}^{2}\right) = \frac{d}{\mathrm{dt}} \left(25\right) R i g h t a r r o w 2 x \frac{\mathrm{dx}}{\mathrm{dt}} + 2 y \frac{\mathrm{dy}}{\mathrm{dt}} = 0$

by dividing by $2 x$,

$R i g h t a r r o w \frac{\mathrm{dx}}{\mathrm{dx}} + \frac{y}{x} \frac{\mathrm{dy}}{\mathrm{dt}} = 0$

by subtracting $\frac{y}{x} \frac{\mathrm{dy}}{\mathrm{dt}}$,

$R i g h t a r r o w \frac{\mathrm{dx}}{\mathrm{dt}} = - \frac{y}{x} \frac{\mathrm{dy}}{\mathrm{dt}}$

Since $y = 4$, $x = \pm 3$, and $\frac{\mathrm{dy}}{\mathrm{dt}} = 6$,

$\frac{\mathrm{dx}}{\mathrm{dt}} = - \frac{4}{\pm 3} \left(6\right) = \pm 8$