Question

a model car travels around a circular track with radius 5 feet. let z denote the distance between the model car and a fixed point that is 20 feet to the left of the center of the circular track. the diagram above indicates the fixed point at the origin, the center of the circular track at the point (20, 0), and the position of the car at the point (x, y). z is the length of the line segment from the origin to the point (x, y). if x and y are functions of time t, in seconds, what is the rate of change of z when x = 23, y = 4, and \\( \frac{dx}{dt}=2 \\)? (the equation of a circle with radius r and center (h, k) is \\( (x - h)^2+(y - k)^2=r^2 \\).) a \\( \frac{dz}{dt}=0 \\), so the distance between the model car and the fixed point is constant. b \\( \frac{dz}{dt}=\frac{40}{\sqrt{545}} \\), so the model car is moving away from the fixed point at a rate of approximately 1.7 feet per second. c \\( \frac{dz}{dt}=\frac{40}{\sqrt{545}} \\), so the model car is moving toward the fixed point at a rate of approximately 1.7 feet per second. d \\( \frac{dz}{dt}=80 \\), so the model car is moving away from the fixed point at a rate of 80

Explanation:

Step1: Use the distance formula

The distance formula \(Z = \sqrt{(x - 0)^2+(y - 0)^2}=\sqrt{x^{2}+y^{2}}\).

Step2: Differentiate \(Z\) with respect to \(t\)

By the chain - rule, \(\frac{dZ}{dt}=\frac{1}{2\sqrt{x^{2}+y^{2}}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})\).
Since the car is on the circle \((x - 20)^2+y^{2}=25\), differentiating both sides with respect to \(t\) gives \(2(x - 20)\frac{dx}{dt}+2y\frac{dy}{dt}=0\). When \(x = 23\) and \(y = 4\), \((23 - 20)\times2+4\times\frac{dy}{dt}=0\), so \(\frac{dy}{dt}=-\frac{3\times2}{4}=-\frac{3}{2}\).

Step3: Substitute values into \(\frac{dZ}{dt}\) formula

Substitute \(x = 23\), \(y = 4\), \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-\frac{3}{2}\) into \(\frac{dZ}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{\sqrt{x^{2}+y^{2}}}\).
\(Z=\sqrt{23^{2}+4^{2}}=\sqrt{529 + 16}=\sqrt{545}\).
\(\frac{dZ}{dt}=\frac{23\times2+4\times(-\frac{3}{2})}{\sqrt{545}}=\frac{46-6}{\sqrt{545}}=\frac{40}{\sqrt{545}}\approx\frac{40}{23.3}\approx1.7\). Since \(\frac{dZ}{dt}>0\), the car is moving away from the fixed point.

Answer:

B. \(\frac{dZ}{dt}=\frac{40}{\sqrt{545}}\), so the model car is moving away from the fixed point at a rate of approximately \(1.7\) feet per second.