Let y=y(x) be a function of x satisfying y√1-x^2=k-x√1-y^2 where k is a constant and y(12)…

Question

Let $y=y(x)$ be a function of $x$ satisfying $y\sqrt{1-x^2}=k-x\sqrt{1-y^2}$ where $k$ is a constant and $y\left(\dfrac12\right)=-\dfrac14$. Then $\dfrac{dy}{dx}$ at $x=\dfrac12$, is equal to:

Accepted Answer

Differentiating implicitly and substituting $x=\dfrac12,\ y=-\dfrac14$ (so $\sqrt{1-x^2}=\dfrac{\sqrt3}2$, $\sqrt{1-y^2}=\dfrac{\sqrt{15}}4$):

$$y'\sqrt{1-x^2}-\frac{xy}{\sqrt{1-y^2}}=-\sqrt{1-y^2}+\frac{xy}{\sqrt{1-x^2}}$$

gives, after substitution, $y'=-\dfrac{\sqrt5}{2}$.

Solutions · 1