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 \\).)

Explanation:

Step1: Find the equation of the circle

The equation of the circle with center \((20,0)\) and radius \(r = 5\) is \((x - 20)^2+y^2=25\).

Step2: Expand the circle equation

Expand \((x - 20)^2+y^2=25\) to get \(x^{2}-40x + 400+y^{2}=25\), or \(x^{2}+y^{2}=40x - 375\).

Step3: Use the distance formula

The distance \(Z=\sqrt{x^{2}+y^{2}}\). Since \(x^{2}+y^{2}=40x - 375\), then \(Z=\sqrt{40x - 375}\).

Step4: Differentiate \(Z\) with respect to \(t\)

Using the chain - rule, \(\frac{dZ}{dt}=\frac{1}{2\sqrt{40x - 375}}\times40\times\frac{dx}{dt}\).
Substitute \(x = 23\) into \(x^{2}+y^{2}=40x - 375\): \(23^{2}+y^{2}=40\times23-375\), \(529 + y^{2}=920 - 375\), \(y^{2}=16\) (which is consistent with \(y = 4\)).
Substitute \(x = 23\) into \(Z=\sqrt{40x - 375}=\sqrt{40\times23-375}=\sqrt{920 - 375}=\sqrt{545}\) (a constant).
Differentiating \(Z\) with respect to \(t\), since \(Z\) is a constant, \(\frac{dZ}{dt}=0\).

Answer:

A. \(\frac{dZ}{dt} = 0\), so the distance between the model car and the fixed point is constant.