A jogger runs around a circular track of radius 45 ft . Let \( (x, y) \) be her coordinates, where the origin is the center of the track. When the jogger's coordinates are \( (27,36) \), her \( x \)-coordinate is changing at a rate of \( 16 \mathrm{ft} / \mathrm{s} \). Find \( d y / d t \). \( d y / d t=\square \mathrm{ft} / \mathrm{s} \) Preview My Answers Submit Answers

To find \( \frac{dy}{dt} \), we start with the relationship given by the equation of the circular track, \( x^2 + y^2 = r^2 \). Since the radius \( r = 45 \) ft, we have: \[ x^2 + y^2 = 45^2 \] At the point \( (27, 36) \): \[ 27^2 + 36^2 = 729 + 1296 = 2025 = 45^2 \] Now, we differentiate both sides of the equation with respect to \( t \): \[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(2025) \] This gives us: \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] Dividing through by 2: \[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \] Now, substitute in the known values \( x = 27 \), \( y = 36 \), and \( \frac{dx}{dt} = 16 \) ft/s: \[ 27(16) + 36 \frac{dy}{dt} = 0 \] Calculating \( 27 \times 16 \): \[ 432 + 36 \frac{dy}{dt} = 0 \] Now, solving for \( \frac{dy}{dt} \): \[ 36 \frac{dy}{dt} = -432 \] \[ \frac{dy}{dt} = \frac{-432}{36} = -12 \] Thus, the rate at which the jogger's \( y \)-coordinate is changing is: \[ \frac{dy}{dt} = -12 \, \text{ft/s} \]