The differentiable functions $x$ and $y$ are related by the following equation:
$$
y^{2}=x^{2}-5
$$

Also, $\frac{d y}{d t}=2.1$.
Find $\frac{d x}{d t}$ when $y=-2$ and $x0$.

Question

The differentiable functions $x$ and $y$ are related by the following equation:
$$
y^{2}=x^{2}-5
$$

Also, $\frac{d y}{d t}=2.1$.
Find $\frac{d x}{d t}$ when $y=-2$ and $x>0$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Differentiate both sides of the equation $y^2 = x^2 - 5$ with respect to $t$ using the chain rule. ‖ ‖ step 2 ⋮ Applying the chain rule gives $2y \frac{dy}{dt} = 2x \frac{dx}{dt}$. ‖ ‖ step 3 ⋮ Solve for $\frac{dx}{dt}$ by dividing both sides by $2x$, yielding $\frac{dx}{dt} = \frac{y}{x} \frac{dy}{dt}$. ‖ ‖ step 4 ⋮ Substitute $y = -2$ and $\frac{dy}{dt} = 2.1$ into the equation from step 3. ‖ ‖ step 5 ⋮ Since $x > 0$ and $y^2 = x^2 - 5$, we find $x$ by solving $(-2)^2 = x^2 - 5$. ‖ ‖ step 6 ⋮ Solving for $x$ gives $x^2 = 4 + 5$, so $x = \sqrt{9} = 3$. ‖ ‖ step 7 ⋮ Substitute $x = 3$ into the equation from step 3 to find $\frac{dx}{dt} = \frac{-2}{3} \cdot 2.1$. ‖ ‖ step 8 ⋮ Calculate $\frac{dx}{dt} = \frac{-2}{3} \cdot 2.1 = -1.4$. ‖ ∻Answer∻ ⚹ $\frac{dx}{dt} = -1.4$ when $y = -2$ and $x > 0$. ⚹ ∻Key Concept∻ ⚹Implicit Differentiation and Chain Rule⚹ ∻Explanation∻ ⚹To find $\frac{dx}{dt}$, we use implicit differentiation on the given equation with respect to $t$, and then solve for $\frac{dx}{dt}$ using the given values for $y$ and $\frac{dy}{dt}$.⚹◊What is the derivative $\frac{dy}{dx}$ with respect to $x$ of the equation $x^{3} y+y^{2}-x^{2}=5$ at the point $(2,1)$?⍭ Generate me a similar question◊

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