BuyFind

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 3.1, Problem 55E
To determine

To find:

The equation of the tangent and parallel to the line

Expert Solution

Answer to Problem 55E

The equation of the both line is y=13(x73)199

Explanation of Solution

Given:

The tangent to the curve

  y=x25x+4

Parallel to the line

  x3y=5y=5x3

Concept used:

The equation is in slope −intercept form, y=mx+c

An equation for the line through the point (x1,y1) with slope m is

  yy1=m(xx1)

Calculation:

The function

  y=x25x+4.......................(1)

The derivative of a function

  y=f(x)y=f(x)=dydx=m

Differentiating the equation (1) with respect to x

  y=x25x+4y=2x5........................(2)

The derivative is slope of the tangent line so in order to the slope of the tangent line

The derivative of constant is zero

  x3y=5y=5x3y=13.............................(3)

The equation is in slope −intercept form

The slope of the line is =13

From equation (2) and equation (3)

  13=2x56x15=1x=146x=73

The coordinates of points

  y=(73)25(73)+4y=499353+4y=559+4y=199

An equation for the line through the point (x1,y1) with slope m is

  yy1=m(xx1)y+199=13(x73)y=13(x73)199

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