Questions: CALCULUS AB SECTION II, Part B Time-45 minutes Number of problems-3 No calculator is allowed for these problems. 4. Consider the curve given by x^2+4 y^2=7+3 x y. (a) Show that dy/dx=(3 y-2 x)/(8 y-3 x). (b) Show that there is a point P with x-coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P. (c) Find the value of d^2 y/dx^2 at the point P found in part (b). Does the curve have a local maximum, a local minimum, or neither at the point P? Justify your answer.

CALCULUS AB
SECTION II, Part B
Time-45 minutes
Number of problems-3
No calculator is allowed for these problems.
4. Consider the curve given by x^2+4 y^2=7+3 x y.
(a) Show that dy/dx=(3 y-2 x)/(8 y-3 x).
(b) Show that there is a point P with x-coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P.
(c) Find the value of d^2 y/dx^2 at the point P found in part (b). Does the curve have a local maximum, a local minimum, or neither at the point P? Justify your answer.

Transcript text: CALCULUS AB SECTION II, Part B Time-45 minutes Number of problems-3 No calculator is allowed for these problems. 4. Consider the curve given by $x^{2}+4 y^{2}=7+3 x y$. (a) Show that $\frac{d y}{d x}=\frac{3 y-2 x}{8 y-3 x}$. (b) Show that there is a point $P$ with $x$-coordinate 3 at which the line tangent to the curve at $P$ is horizontal. Find the $y$-coordinate of $P$. (c) Find the value of $\frac{d^{2} y}{d x^{2}}$ at the point $P$ found in part (b). Does the curve have a local maximum, a local minimum, or neither at the point $P$ ? Justify your answer.

Solution

Solution Steps

Solution Approach

Part (a)

Start with the given equation $x^2 + 4y^2 = 7 + 3xy$.
Differentiate both sides implicitly with respect to $x$.
Solve for $\frac{dy}{dx}$ to show that $\frac{dy}{dx} = \frac{3y - 2x}{8y - 3x}$.

Part (b)

Use the result from part (a) to find the condition for the tangent line to be horizontal, which is when $\frac{dy}{dx} = 0$.
Set the numerator of $\frac{dy}{dx}$ to zero and solve for $y$ when $x = 3$.

Part (c)

Use the second derivative test to find $\frac{d^2y}{dx^2}$ at the point $P$.
Determine whether the curve has a local maximum, minimum, or neither at point $P$ by analyzing the sign of $\frac{d^2y}{dx^2}$.

Step 1: Implicit Differentiation to Find $\frac{dy}{dx}$

Given the curve: \[ x^2 + 4y^2 = 7 + 3xy \]

Differentiate both sides with respect to $x$: \[ \frac{d}{dx}(x^2) + \frac{d}{dx}(4y^2) = \frac{d}{dx}(7) + \frac{d}{dx}(3xy) \]

Using the chain rule and product rule: \[ 2x + 4 \cdot 2y \frac{dy}{dx} = 0 + 3 \left( y + x \frac{dy}{dx} \right) \]

Simplify: \[ 2x + 8y \frac{dy}{dx} = 3y + 3x \frac{dy}{dx} \]

Rearrange to solve for $\frac{dy}{dx}$: \[ 8y \frac{dy}{dx} - 3x \frac{dy}{dx} = 3y - 2x \]

Factor out $\frac{dy}{dx}$: \[ (8y - 3x) \frac{dy}{dx} = 3y - 2x \]

Divide both sides by $8y - 3x$: \[ \frac{dy}{dx} = \frac{3y - 2x}{8y - 3x} \]

Final Answer for Part (a)

\[ \boxed{\frac{dy}{dx} = \frac{3y - 2x}{8y - 3x}} \]

Step 2: Finding the Point $P$ Where the Tangent Line is Horizontal

For the tangent line to be horizontal, $\frac{dy}{dx} = 0$: \[ \frac{3y - 2x}{8y - 3x} = 0 \]

This implies: \[ 3y - 2x = 0 \] \[ 3y = 2x \] \[ y = \frac{2}{3}x \]

Given $x = 3$: \[ y = \frac{2}{3} \cdot 3 = 2 \]

So, the point $P$ is $(3, 2)$.

Final Answer for Part (b)

\[ \boxed{(3, 2)} \]

Step 3: Finding $\frac{d^2y}{dx^2}$ at Point $P$ and Determining Local Extrema

First, we need to find $\frac{d^2y}{dx^2}$. Differentiate $\frac{dy}{dx} = \frac{3y - 2x}{8y - 3x}$ implicitly with respect to $x$:

Let $u = 3y - 2x$ and $v = 8y - 3x$: \[ \frac{dy}{dx} = \frac{u}{v} \]

Using the quotient rule: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculate $\frac{du}{dx}$ and $\frac{dv}{dx}$: \[ \frac{du}{dx} = 3 \frac{dy}{dx} - 2 \] \[ \frac{dv}{dx} = 8 \frac{dy}{dx} - 3 \]

Substitute $\frac{dy}{dx} = \frac{3y - 2x}{8y - 3x}$ at $(3, 2)$: \[ \frac{dy}{dx} \bigg|_{(3, 2)} = \frac{3(2) - 2(3)}{8(2) - 3(3)} = \frac{6 - 6}{16 - 9} = 0 \]

So: \[ \frac{du}{dx} = 3(0) - 2 = -2 \] \[ \frac{dv}{dx} = 8(0) - 3 = -3 \]

Now, substitute back into the quotient rule: \[ \frac{d^2y}{dx^2} = \frac{v(-2) - u(-3)}{v^2} \]

At $(3, 2)$: \[ u = 3(2) - 2(3) = 0 \] \[ v = 8(2) - 3(3) = 16 - 9 = 7 \]

So: \[ \frac{d^2y}{dx^2} = \frac{7(-2) - 0(-3)}{7^2} = \frac{-14}{49} = -\frac{2}{7} \]

Since $\frac{d^2y}{dx^2} < 0$, the curve has a local maximum at point $P$.

Final Answer for Part (c)

\[ \boxed{\frac{d^2y}{dx^2} = -\frac{2}{7}, \text{ local maximum}} \]

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