problem 10 : find an equation of the line tangent to the graph of $(x^{2}+y^{2})^{3}=8x^{2}y^{2}$ at the…

Explanation:

Step1: Verify point on curve

Substitute $x=-1, y=1$ into $(x^2+y^2)^3 = 8x^2y^2$: $$((-1)^2+1^2)^3=(1+1)^3=8$$ $$8(-1)^2(1)^2=8\times1\times1=8$$ Both sides equal, so the point is on the curve.

Step2: Implicit differentiation

Differentiate both sides w.r.t. $x$: $$3(x^2+y^2)^2\cdot(2x+2y\frac{dy}{dx}) = 8\left(2x y^2 + x^2\cdot2y\frac{dy}{dx}\right)$$

Step3: Simplify the equation

Factor out constants and simplify: $$6(x^2+y^2)^2\left(x+y\frac{dy}{dx}\right) = 16xy\left(y + x\frac{dy}{dx}\right)$$ Divide both sides by 2: $$3(x^2+y^2)^2\left(x+y\frac{dy}{dx}\right) = 8xy\left(y + x\frac{dy}{dx}\right)$$

Step4: Substitute $x=-1,y=1$

First calculate $x^2+y^2=(-1)^2+1^2=2$: $$3(2)^2\left(-1+1\cdot\frac{dy}{dx}\right) = 8(-1)(1)\left(1 + (-1)\frac{dy}{dx}\right)$$ $$12\left(-1+\frac{dy}{dx}\right) = -8\left(1-\frac{dy}{dx}\right)$$

Step5: Solve for $\frac{dy}{dx}$

Expand both sides: $$-12 + 12\frac{dy}{dx} = -8 + 8\frac{dy}{dx}$$ Rearrange terms: $$12\frac{dy}{dx}-8\frac{dy}{dx} = -8 + 12$$ $$4\frac{dy}{dx}=4$$ $$\frac{dy}{dx}=1$$

Step6: Write tangent line equation

Use point-slope form $y-y_1=m(x-x_1)$ with $m=1, (x_1,y_1)=(-1,1)$: $$y-1=1\cdot(x-(-1))$$ Simplify: $$y-1=x+1$$

Answer:

$y = x + 2$