Question

Math

Posted 4 months ago

```
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 $x>0$.
```

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Answer from Sia

Posted 4 months ago

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 $\frac{dy}{dt}$.

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