QuestionSet $x^{*}=1$. For each of the implicit curves below, do the following: - Find $y^{*}$ (there may be more than one possible $y^{*}$ value) where $\left(x^{*}, y^{*}\right)$ is a point on the curve and - calculate $\left.\frac{d y}{d x}\right|_{\left(x^{*}, y^{*}\right)}$.
The implicit curves are (a) $x^{2} y+6 x^{4} y^{2}=\sin (\pi x)$ (b) $\frac{(x-1) y}{\sqrt{1+x^{2}}}+(1+x y)=\tan ((x-1) y)$ (c) $x^{2}-y^{2}=\cos \left(\frac{\pi}{2} x+(x-1) y\right)$

Studdy Solution
Evaluate $\frac{dy}{dx}$ at $(x^{*}, y^{*}) = (1, 0)$ and $(1, -\frac{1}{6})$.
For $y^{*} = 0$:
$$\frac{dy}{dx} = \frac{\pi \cos(\pi \cdot 1) - 2 \cdot 1 \cdot 0 - 24 \cdot 1^3 \cdot 0^2}{1^2 + 12 \cdot 1^4 \cdot 0}$$ $$\frac{dy}{dx} = \frac{\pi \cdot (-1)}{1} = -\pi$$
For $y^{*} = -\frac{1}{6}$:
$$\frac{dy}{dx} = \frac{\pi \cos(\pi \cdot 1) - 2 \cdot 1 \cdot (-\frac{1}{6}) - 24 \cdot 1^3 \cdot \left(-\frac{1}{6}\right)^2}{1^2 + 12 \cdot 1^4 \cdot (-\frac{1}{6})}$$
Calculate values:
$$\frac{dy}{dx} = \frac{-\pi + \frac{1}{3} - \frac{24}{36}}{1 - 2}$$ $$\frac{dy}{dx} = \frac{-\pi + \frac{1}{3} - \frac{2}{3}}{-1}$$ $$\frac{dy}{dx} = \pi - \frac{1}{3} + \frac{2}{3}$$ $$\frac{dy}{dx} = \pi + \frac{1}{3}$$
(b) Curve: $\frac{(x-1) y}{\sqrt{1+x^{2}}}+(1+x y)=\tan ((x-1) y)$
STEP_1: Substitute $x^{*} = 1$ into the equation:
$$\frac{(1-1) y}{\sqrt{1+1^{2}}}+(1+1 \cdot y)=\tan ((1-1) y)$$ $$0 + (1 + y) = \tan(0)$$ $$1 + y = 0$$
Thus, $y = -1$.
STEP_2: Differentiate the equation implicitly with respect to $x$:
$$\frac{d}{dx}\left(\frac{(x-1) y}{\sqrt{1+x^{2}}}\right) + \frac{d}{dx}(1 + xy) = \frac{d}{dx}(\tan((x-1)y))$$
Using the product rule, quotient rule, and chain rule:
$$\frac{(x-1) \frac{dy}{dx}}{\sqrt{1+x^2}} + \frac{y}{\sqrt{1+x^2}} - \frac{(x-1)y x}{(1+x^2)^{3/2}} + y + x\frac{dy}{dx} = \sec^2((x-1)y)((x-1)\frac{dy}{dx} + y)$$
Rearrange to solve for $\frac{dy}{dx}$:
$$\frac{dy}{dx} \left(\frac{x-1}{\sqrt{1+x^2}} + x - (x-1)\sec^2((x-1)y)\right) = \sec^2((x-1)y)y - \frac{y}{\sqrt{1+x^2}} + \frac{(x-1)yx}{(1+x^2)^{3/2}} - y$$
$$\frac{dy}{dx} = \frac{\sec^2((x-1)y)y - \frac{y}{\sqrt{1+x^2}} + \frac{(x-1)yx}{(1+x^2)^{3/2}} - y}{\frac{x-1}{\sqrt{1+x^2}} + x - (x-1)\sec^2((x-1)y)}$$
STEP_3: Evaluate $\frac{dy}{dx}$ at $(x^{*}, y^{*}) = (1, -1)$.
$$\frac{dy}{dx} = \frac{\sec^2(0)(-1) - \frac{-1}{\sqrt{2}} + \frac{0}{2^{3/2}} + 1}{\frac{0}{\sqrt{2}} + 1 - 0}$$
Calculate values:
$$\frac{dy}{dx} = \frac{-1 + \frac{1}{\sqrt{2}} + 1}{1}$$ $$\frac{dy}{dx} = \frac{1}{\sqrt{2}}$$
(c) Curve: $x^{2} - y^{2} = \cos \left(\frac{\pi}{2} x + (x-1) y\right)$
STEP_1: Substitute $x^{*} = 1$ into the equation:
$$1^2 - y^2 = \cos \left(\frac{\pi}{2} \cdot 1 + (1-1) y\right)$$ $$1 - y^2 = \cos\left(\frac{\pi}{2}\right)$$ $$1 - y^2 = 0$$
Thus, $y^2 = 1$ and $y = \pm 1$.
STEP_2: Differentiate the equation implicitly with respect to $x$:
$$\frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx}\left(\cos\left(\frac{\pi}{2} x + (x-1)y\right)\right)$$
Using the chain rule:
$$2x - 2y \frac{dy}{dx} = -\sin\left(\frac{\pi}{2} x + (x-1)y\right)\left(\frac{\pi}{2} + (x-1)\frac{dy}{dx} + y\right)$$
Rearrange to solve for $\frac{dy}{dx}$:
$$2y \frac{dy}{dx} + \sin\left(\frac{\pi}{2} x + (x-1)y\right)(x-1)\frac{dy}{dx} = 2x + \sin\left(\frac{\pi}{2} x + (x-1)y\right)\left(\frac{\pi}{2} + y\right)$$
$$\frac{dy}{dx} (2y + \sin\left(\frac{\pi}{2} x + (x-1)y\right)(x-1)) = 2x + \sin\left(\frac{\pi}{2} x + (x-1)y\right)\left(\frac{\pi}{2} + y\right)$$
$$\frac{dy}{dx} = \frac{2x + \sin\left(\frac{\pi}{2} x + (x-1)y\right)\left(\frac{\pi}{2} + y\right)}{2y + \sin\left(\frac{\pi}{2} x + (x-1)y\right)(x-1)}$$
STEP_3: Evaluate $\frac{dy}{dx}$ at $(x^{*}, y^{*}) = (1, 1)$ and $(1, -1)$.
For $y^{*} = 1$:
$$\frac{dy}{dx} = \frac{2 \cdot 1 + \sin\left(\frac{\pi}{2} \cdot 1 + 0\right)\left(\frac{\pi}{2} + 1\right)}{2 \cdot 1 + \sin\left(\frac{\pi}{2} \cdot 1 + 0\right) \cdot 0}$$
Calculate values:
$$\frac{dy}{dx} = \frac{2 + 1 \cdot \left(\frac{\pi}{2} + 1\right)}{2}$$ $$\frac{dy}{dx} = \frac{2 + \frac{\pi}{2} + 1}{2}$$ $$\frac{dy}{dx} = \frac{3 + \frac{\pi}{2}}{2}$$
For $y^{*} = -1$:
$$\frac{dy}{dx} = \frac{2 \cdot 1 + \sin\left(\frac{\pi}{2} \cdot 1 + 0\right)\left(\frac{\pi}{2} - 1\right)}{2 \cdot (-1) + \sin\left(\frac{\pi}{2} \cdot 1 + 0\right) \cdot 0}$$
Calculate values:
$$\frac{dy}{dx} = \frac{2 + 1 \cdot \left(\frac{\pi}{2} - 1\right)}{-2}$$ $$\frac{dy}{dx} = \frac{2 + \frac{\pi}{2} - 1}{-2}$$ $$\frac{dy}{dx} = \frac{1 + \frac{\pi}{2}}{-2}$$
The solutions for each curve are:
(a) $y^{*} = 0$, $\frac{dy}{dx} = -\pi$; $y^{*} = -\frac{1}{6}$, $\frac{dy}{dx} = \pi + \frac{1}{3}$
(b) $y^{*} = -1$, $\frac{dy}{dx} = \frac{1}{\sqrt{2}}$
(c) $y^{*} = 1$, $\frac{dy}{dx} = \frac{3 + \frac{\pi}{2}}{2}$; $y^{*} = -1$, $\frac{dy}{dx} = \frac{1 + \frac{\pi}{2}}{-2}$