Question

Assume that \(x\) and \(y\) are both differentiable functions of \(t\) and find the required values of \(d y / d t\) and \(d x / d t\) $$ \begin{aligned} &x^{2}+y^{2}=25 \quad \text { (a) } \frac{d y}{d t} \text { when } x=3, y=4 \quad \frac{d x}{d t}=8\\\ &\text { (b) } \frac{d x}{d t} \text { when } x=4, y=3 \quad \frac{d y}{d t}=-2 \end{aligned} $$

Short answer

The required value of \( \frac{dy}{dt} \) in part (a) is -6. The required value of \( \frac{dx}{dt} \) in part (b) is 1.5.

Step-by-step solution

Differentiate the equation with respect to \( t \)

Starting with equation \( x^2 + y^2 = 25 \), applying the chain rule, we differentiate both sides with respect to \( t \) to get: \( 2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0 \)

Solve for \(dy/dt\) in part (a)

Isolate \(dy/dt\) to get: \( \frac{dy}{dt} = -\frac{x}{y} \cdot \frac{dx}{dt} \). Subsitute (x=3, y=4, \( \frac{dx}{dt} = 8 \)) into the equation to get: \( \frac{dy}{dt} = -\frac{3}{4} \cdot 8 = -6 \)

Solve for \(dx/dt\) in part (b)

Isolate \(dx/dt\) to get: \( \frac{dx}{dt} = -\frac{y}{x} \cdot \frac{dy}{dt} \). Subsitute (x=4, y=3, \( \frac{dy}{dt} = -2 \)) into the equation to get: \( \frac{dx}{dt} = -\frac{3}{4} \cdot -2 = 1.5 \)