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Consider the circle of radius 1 centered at the origin, that is, the solutions of the equation

x2+y2=1

(a) Find all points on the graph with an x-coordinate of x = 1/2, and then find the slope of the tangent line at each of these points.

(b) Find all points on the graph with a y-coordinate of y=22, and then find the slope of the tangent line at each of these points.

(c) Find all points on the graph where the tangent line is vertical.

(d) Find all points on the graph where the tangent line has a slope of −1.

Short Answer

Expert verified

Part (a): 12,-3212,32arethepointsonthegraph.dydx=13&-13

Part (b): 22,22-22,22arethepointsonthegraph.dydx=1&-1

Part (c): 1,0-1,0arethepointsonthegraph.

Part (d): 12,12-12,-12arethepointsonthegraph.

Step by step solution

01

Step 1. Given information is:

Functionofthecircleis:x2+y2=1

02

Part (a) Step 1. Calculating dy/dx

Here,x2+y2=1dx2+y2dx=0(Differentiatingbothsides)2x+2ydydx=02ydydx=-2xdydx=-2x2ydydx=-xy

03

Part (a) Step 2. Calculating y and slope 

Substitutingx=12intotheequationx2+y2=1:122+y2=1y2=34y=±32Thus,12,-3212,32arethepointsonthegraph.Thustheslopeis:i)whenx=12,y=-32dydx=-12-32dydx=13ii)whenx=12,y=32dydx=-1232dydx=-13

04

Part (b) Step 1. Calculating x and slope

Substitutingy=22intotheequationx2+y2=1:x2+222=1x2=24x=±22Thus,22,22-22,22arethepointsonthegraph.Thustheslopeis:i)whenx=22,y=22dydx=-2222dydx=-1ii)whenx=-22,y=22dydx=-2222dydx=1

05

Part (c) Step 1. Finding points

Substitutingy=0intotheequationx2+y2=1forthetangentlinetobevertical:x2+02=1x2=1x=±1Thus,1,0-1,0arethepointsonthegraph.

06

Part (d) Step 1. Finding points

Here,-xy=-1x=ySubstitutingtheaboverelationingivenequation:y2+y2=12y2=1y2=12y=±12Thus,x2+122=1x2+12=1x=±12Thus,12,12-12,-12arethepointsonthegraph.

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Most popular questions from this chapter

Use the definition of the derivative to find the equations of the lines described in Exercises 59-64.

The tangent line to f(x)=x2 atx=0

Every morning Linda takes a thirty-minute jog in Central Park. Suppose her distance s in feet from the oak tree on the north side of the park tminutes after she begins her jog is given by the function s(t)shown that follows at the left, and suppose she jogs on a straight path leading into the park from the oak tree.

(a) What was the average rate of change of Linda’s distance from the oak tree over the entire thirty-minute jog? What does this mean in real-world terms?

(b) On which ten-minute interval was the average rate of change of Linda’s distance from the oak tree the greatest: the first 10minutes, the second 10minutes, or the last10minutes?

(c) Use the graph of s(t)to estimate Linda’s average velocity during the 5-minute interval fromt=5tot=10. What does the sign of this average velocity tell you in real-world terms?

(d) Approximate the times at which Linda’s (instantaneous) velocity was equal to zero. What is the physical significance of these times?

(e) Approximate the time intervals during Linda’s jog that her (instantaneous) velocity was negative. What does a negative velocity mean in terms of this physical example?

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