Assume that \( x \) and \( y \) are both differentiable functions of \( t \). Find the required values of \( \frac{d y}{d t} \) and \( \frac{d x}{d t} \). \[ x^{2}+y^{2}=169 \] (a) Find \( \frac{d y}{d t} \), given \( x=5, y=12 \), and \( \frac{d x}{d t}=7 \). \[ \frac{d y}{d t}= \] \( \square \) (b) Find \( \frac{d x}{d t} \), given \( x=12, y=5 \), and \( \frac{d y}{d t}=-3 \). \[ \frac{d x}{d t}=\square \]

To solve the given problem, we start with the equation of the circle: \[ x^2 + y^2 = 169 \] Now, we will differentiate both sides of this equation with respect to \( t \): \[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(169) \] Using the chain rule, we find: \[ 2x \frac{d x}{d t} + 2y \frac{d y}{d t} = 0 \] We can simplify this to: \[ x \frac{d x}{d t} + y \frac{d y}{d t} = 0 \] ### (a) Finding \( \frac{d y}{d t} \) Given \( x = 5 \), \( y = 12 \), and \( \frac{d x}{d t} = 7 \): Substituting these values into our derived equation gives: \[ 5 \cdot 7 + 12 \frac{d y}{d t} = 0 \] \[ 35 + 12 \frac{d y}{d t} = 0 \] \[ 12 \frac{d y}{d t} = -35 \] \[ \frac{d y}{d t} = -\frac{35}{12} \] So, the final answer for (a) is: \[ \frac{d y}{d t} = -\frac{35}{12} \, \square \] ### (b) Finding \( \frac{d x}{d t} \) Given \( x = 12 \), \( y = 5 \), and \( \frac{d y}{d t} = -3 \): Substituting these values into our derived equation gives: \[ 12 \frac{d x}{d t} + 5 \cdot (-3) = 0 \] \[ 12 \frac{d x}{d t} - 15 = 0 \] \[ 12 \frac{d x}{d t} = 15 \] \[ \frac{d x}{d t} = \frac{15}{12} = \frac{5}{4} \] So, the final answer for (b) is: \[ \frac{d x}{d t} = \frac{5}{4} \, \square \]