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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 dxdt=2 ?

Respuesta :

Answer:

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

Step-by-step explanation:

The functions x and y satisfy (x−20)^2+y^2=25 and differentiating gives 2(x−20)dx/dt+2y dy/dt=0. Substituting the three known values and solving for dy/dt yields dy/dt=−32. Since Z=x^2+y^2−−−−−−√, dZ/dt=(2x ^ dx/dt+2y^dy/dt) / 2x2+y2√. Substituting Since Substituting for x, y,[tex]\frac{dx}{dt}[/tex] , and [tex]\frac{dy}{dt}[/tex] gives [tex]\frac{dZ}{dt}[/tex]= [tex]\frac{40}{\sqrt{545} }[/tex] ,

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