QuestionIf $x^{2}+y^{2}=13$ and $\frac{\mathrm{d} y}{\mathrm{~d} t}=8$, find $\frac{\mathrm{d} x}{\mathrm{~d} t}$ when $y=3$ and $x=2$. $$\left.\frac{\mathrm{d} x}{\mathrm{~d} t}\right|_{(x, y)=(2,3)}=$$

Studdy Solution
Solve for $\frac{\mathrm{d} x}{\mathrm{~d} t}$:
$4 \frac{\mathrm{d} x}{\mathrm{~d} t} = -48$
Divide both sides by 4:
$\frac{\mathrm{d} x}{\mathrm{~d} t} = -12$
The value of $\left.\frac{\mathrm{d} x}{\mathrm{~d} t}\right|_{(x, y)=(2,3)}$ is:
$\boxed{-12}$