1. 2008 free - response (form b) #6\n6. consider the closed curve in the xy - plane given by (x^{2}+2x +…

1. 2008 free - response (form b) #6\n6. consider the closed curve in the xy - plane given by (x^{2}+2x + y^{4}+4y = 5).\n(a) show that (\frac{dy}{dx}=\frac{-(x + 1)}{2(y^{3}+1)}).\n(b) write an equation for the line tangent to the curve at the point (-2,1).\n(c) find the coordinates of the two points on the curve where the line tangent to the curve is vertical.\n(d) is it possible for this curve to have a horizontal tangent at points where it intersects the x - axis? explain your reasoning.\n(e) evaluate (\frac{d^{2}y}{dx^{2}}) at the point on the curve where (x=-2) and (y = 1).

1. 2008 free - response (form b) #6\n6. consider the closed curve in the xy - plane given by (x^{2}+2x + y^{4}+4y = 5).\n(a) show that (\frac{dy}{dx}=\frac{-(x + 1)}{2(y^{3}+1)}).\n(b) write an equation for the line tangent to the curve at the point (-2,1).\n(c) find the coordinates of the two points on the curve where the line tangent to the curve is vertical.\n(d) is it possible for this curve to have a horizontal tangent at points where it intersects the x - axis? explain your reasoning.\n(e) evaluate (\frac{d^{2}y}{dx^{2}}) at the point on the curve where (x=-2) and (y = 1).

Answer

Explanation:

Step1: Differentiate implicitly

Differentiate $x^{2}+2x + y^{4}+4y = 5$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ and the chain - rule $\frac{d}{dx}(y^{n})=ny^{n - 1}\frac{dy}{dx}$, we get: $2x + 2+4y^{3}\frac{dy}{dx}+4\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(4y^{3}+4)=-(2x + 2)$. Then $\frac{dy}{dx}=\frac{-(x + 1)}{2(y^{3}+1)}$.

Step3: Find the slope of the tangent line at $(-2,1)$

Substitute $x=-2$ and $y = 1$ into $\frac{dy}{dx}$: $\frac{dy}{dx}\big|{x=-2,y = 1}=\frac{-(-2 + 1)}{2(1^{3}+1)}=\frac{1}{4}$. Using the point - slope form of a line $y - y{1}=m(x - x_{1})$ where $(x_{1},y_{1})=(-2,1)$ and $m=\frac{1}{4}$, the equation of the tangent line is $y - 1=\frac{1}{4}(x + 2)$, or $y=\frac{1}{4}x+\frac{3}{2}$.

Step4: Find points with vertical tangent

A vertical tangent occurs when $\frac{dy}{dx}$ is undefined, i.e., when $2(y^{3}+1)=0$. $y^{3}+1 = 0$ gives $y=-1$. Substitute $y = - 1$ into the original equation $x^{2}+2x+(-1)^{4}+4(-1)=5$. $x^{2}+2x+1 - 4 = 5$, $x^{2}+2x-8 = 0$. Factor: $(x + 4)(x - 2)=0$. So $x=-4$ or $x = 2$. The points are $(-4,-1)$ and $(2,-1)$.

Step5: Analyze horizontal tangents at $x -$axis intersections

The curve intersects the $x$ - axis when $y = 0$. The original equation becomes $x^{2}+2x=5$, $x^{2}+2x - 5=0$. Using the quadratic formula $x=\frac{-2\pm\sqrt{4+20}}{2}=-1\pm\sqrt{6}$. A horizontal tangent occurs when $\frac{dy}{dx}=0$, i.e., $x+1 = 0$ or $x=-1$. Since $x=-1\neq-1\pm\sqrt{6}$, it is not possible for the curve to have a horizontal tangent at points where it intersects the $x$ - axis.

Step6: Find the second - derivative

We use the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$ where $u=-(x + 1)$ and $v=2(y^{3}+1)$. $u'=-1$ and $v'=6y^{2}\frac{dy}{dx}$. $\frac{d^{2}y}{dx^{2}}=\frac{-1\times2(y^{3}+1)-[-(x + 1)]\times6y^{2}\frac{dy}{dx}}{4(y^{3}+1)^{2}}$. Substitute $x=-2$, $y = 1$ and $\frac{dy}{dx}=\frac{1}{4}$: $\frac{d^{2}y}{dx^{2}}=\frac{-2(1^{3}+1)-[-(-2 + 1)]\times6\times1^{2}\times\frac{1}{4}}{4(1^{3}+1)^{2}}$ $=\frac{-4-(1)\times\frac{3}{2}}{16}=\frac{-\frac{8 + 3}{2}}{16}=-\frac{11}{32}$.

Answer:

(a) Shown above. (b) $y=\frac{1}{4}x+\frac{3}{2}$ (c) $(-4,-1)$ and $(2,-1)$ (d) No. When $y = 0$, $x=-1\pm\sqrt{6}$, and for horizontal tangent $x=-1$. (e) $-\frac{11}{32}$