If 3x^2y+2x=-32 and dy/dt=-4 when x=2 and y=-3, how do you find dx/dt?

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If 3x^2y+2x=-32 and dy/dt=-4 when x=2 and y=-3, how do you find dx/dt?

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Answer 1 · 2018-01-15T11:52:34

The chain rule implies $\frac{d x}{d t} = \frac{\frac{d y}{d t}}{\frac{d y}{d x}} [1]$ $dx/dt = (dy/dt)/(dy/dx)" [1]"$
Use implicit differentiation to obtain $\frac{d y}{d x}$ $dy/dx$
Evaluate $\frac{d y}{d x}$ $dy/dx$ at $(2, - 3)$ $(2,-3)$
Substitute the result and $\frac{d y}{d t}$ $dy/dt$ into equation [1]

Explanation:

Given: $3 x^{2} y + 2 x = - 32$ $3x^2y+2x=-32$

Verify that the point $(2, - 3)$ $(2,-3)$ lies on the curve:

$3 {(2)}^{2} (- 3) + 2 (2) = - 32$ $3(2)^2(-3)+2(2)=-32$

Use implicit differentiation:

$6 x y + 3 x^{2} \frac{d y}{d x} + 2 = 0$ $6xy+3x^2dy/dx + 2 = 0$

$3 x^{2} \frac{d y}{d x} = - 2 - 6 x y$ $3x^2dy/dx = -2-6xy$

$\frac{d y}{d x} = \frac{- 2 - 6 x y}{3 x^{2}}$ $dy/dx = (-2-6xy)/(3x^2)$

Evaluate $\frac{d y}{d x}$ $dy/dx$ at $(2, - 3)$ $(2,-3)$

$\frac{d y}{d x} = \frac{- 2 - 6 (2) (- 3)}{3 {(2)}^{2}}$ $dy/dx = (-2-6(2)(-3))/(3(2)^2)$

$\frac{d y}{d x} = \frac{17}{6}$ $dy/dx = 17/6$

Substitute $\frac{d y}{d x} = \frac{17}{6}$ $dy/dx = 17/6$ and $\frac{d y}{d t} = - 4$ $dy/dt = -4$ into equation [1]:

$\frac{d x}{d t} = \frac{- 4}{\frac{17}{6}}$ $dx/dt = (-4)/(17/6)$

$\frac{d x}{d t} = - \frac{24}{17}$ $dx/dt = -24/17$

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If 3x^2y+2x=-32 and dy/dt=-4 when x=2 and y=-3, how do you find dx/dt?

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Explanation:

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