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Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.4, Problem 85E

a.

To determine

Explain that the given curve has two tangent lines at the given point.

Expert Solution

Answer to Problem 85E

The equation of tangent lines are y=3x+33 and y=3x33 .

Explanation of Solution

Given:

The given equations are x=t2 and y=t33t . The given point is (3,0) .

Calculation:

Find the slope.

Apply formula dydx=dydtdxdt .

  dydt=ddt(t33t)

Apply differencerule (fg)'=f'g' .

  dydt=ddt(t3)ddt(3t)

Use derivative rule ddx(xn)=nxn1 .

  dydt=3t23dydt=3(t21)

  dxdt=ddt(t2)

Use derivative rule ddx(xn)=nxn1 .

  dxdt=2t

  dydx(slope)=3(t21)2t

Find the value of the parameter t at the point (3,0) .

At x=3

  t2=3t=±3

At y=0

  t33t=0t(t23)=0t=0andt23=0t2=3t=±3

Now,

  t=±3 , because it satisfies both the equations.

Slope at t=3 .

  dydx(slope)=3(t21)2t=3[(3)21]2(3)=3[31]2(3)=33=33×33=3

Slope at t=3 .

  dydx(slope)=3(t21)2t=3[(3)21]2(3)=3[31]2(3)=33=33×33=3

There are two different tangents at the given point, because there is two different slope values.

Now use point-slope form for the tangent equations.

  yy1=m(xx1)y1=0,x1=3,m=3y0=3(x3)y=3x+33andy1=0,x1=3,m=3y0=3(x3)y=3x33

Hence the equation of tangent lines are y=3x+33 and y=3x33 .

b.

To determine

Find the horizontal and vertical tangent line points on the curve.

Expert Solution

Answer to Problem 85E

The tangents are horizontal at the points (2,2) and (2,2) . The tangent is vertical at the point (0,0) .

Explanation of Solution

Given:

The given equations are x=t2 and y=t33t . The given point is (3,0) .

Calculation:

Find the slope.

Apply formula dydx=dydtdxdt .

  dydt=ddt(t33t)

Apply difference rule (fg)'=f'g' .

  dydt=ddt(t3)ddt(3t)

Use derivative rule ddx(xn)=nxn1 .

  dydt=3t23dydt=3(t21)

  dxdt=ddt(t2)

Use derivative rule ddx(xn)=nxn1 .

  dxdt=2t

  dydx(slope)=3(t21)2t

For the horizontal tangent line.

  dydx=03(t21)2t=03(t21)=0t21=0t2=1t=±1t=±1

Substitute t=1 in the given equations.

  x=t2=(1)2=1y=t33t=(1)33(1)=2(x,y)=(1,2)

Substitute t=1 in the given equations.

  x=t2=(1)2=1y=t33t=(1)33(1)=2(x,y)=(1,2)

For vertical tangent line dxdt=0 .

  dxdt=02t=0t=0

Substitute t=0 in the given equations.

  x=t2=02=0y=t33t=(0)33(0)=0(x,y)=(0,0)

Hence the tangents are horizontal at the points (2,2) and (2,2) . The tangent is vertical at the point (0,0) .

c.

To determine

Draw a graph for the part (a) and part (b) .

Expert Solution

Explanation of Solution

Given:

The given equations are x=t2 and y=t33t . The given point is (3,0) .

Calculation:

Find the slope.

Apply formula dydx=dydtdxdt .

  dydt=ddt(t33t)

Apply difference rule (fg)'=f'g' .

  dydt=ddt(t3)ddt(3t)

Use derivative rule ddx(xn)=nxn1 .

  dydt=3t23dydt=3(t21)

  dxdt=ddt(t2)

Use derivative rule ddx(xn)=nxn1 .

  dxdt=2t

  dydx(slope)=3(t21)2t

For the horizontal tangent line.

  dydx=03(t21)2t=03(t21)=0t21=0t2=1t=±1t=±1

Substitute t=1 in the given equations.

  x=t2=(1)2=1y=t33t=(1)33(1)=2(x,y)=(1,2)

Substitute t=1 in the given equations.

  x=t2=(1)2=1y=t33t=(1)33(1)=2(x,y)=(1,2)

For vertical tangent line dxdt=0 .

  dxdt=02t=0t=0

Substitute t=0 in the given equations.

  x=t2=02=0y=t33t=(0)33(0)=0(x,y)=(0,0)

Draw a table for the curve C .

    tx=t2y=t33t
    012340149160221852

Hence the graph of the curve is given below.

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.4, Problem 85E

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