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 3.4, Problem 50E
To determine

To find: The point on which the curve is the tangent line perpendicular to the given line.

Expert Solution

Answer to Problem 50E

The point on which the curve is the tangent line perpendicular to the given line is (4, 3).

Explanation of Solution

Given:

The equation of the curve y=1+2x.

The equation of the line 6x+2y=1 .

Result used: Chain Rule

If h is differentiable at x and g is differentiable at h(x), then the composite function F=gh defined by F(x)=g(h(x)) is differentiable at x and F is given by the product

F(x)=g(h(x))h(x) (1)

Derivative rules:

(1) Constant Multiple Rule: ddx[cf(x)]=cddxf(x)

(2) Power Rule: ddx(xn)=nxn1

Calculation:

The derivative of y is dydx, which is obtained as follows,

dydx=ddx(y) =ddx(1+2x) 

Let h(x)=1+2x and g(u)=u  where u=h(x).

Apply the chain rule as shown in equation (1),

dydx=g(h(x))h(x) (2)

The derivative of g(h(x)) is computed as follows,

g(h(x))=g(u)=ddu(g(u))=ddu(u)=ddu(u)12

Apply the power rule (2),

g(h(x))=12(u)121=12(u)122=12(u)12=12u

Substitute u=1+2x in the above equation,

g(h(x))=121+2x

Thus, the derivative g(h(x)) is g(h(x))=121+2x.

The derivative of h(x) is computed as follows,

h(x)=ddx(1+2x)=ddx(1)+ddx(2x)

Apply the constant multiple rule (1),

h(x)=ddx(1)+2ddx(x)=0+2(1x11)=2 

Thus, the derivative h(x) is h(x)=2.

Substitute 121+2x for g(h(x)) and 2 for h(x) in equation (2),

dydx=121+2x2=221+2x=11+2x

Therefore, the derivative of the curve y is dydx=11+2x .

Obtain the slope of the line 6x+2y=1 as follows,

Rewrite the line equation as slope intercept form.

6x+2y=12y=6x+1y=62x+12y=3x+12

Therefore, the slope of the line is m1=3.

Obtain the point on the curve is if slope of tangent line is perpendicular to slope of the line 6x+2y=1.

Note that, if two lines are perpendicular with slopes are m1 and m2, then the product of their slopes is m1m2=1.

The required slope (m2) of the tangent line is perpendicular to the given line 6x+2y=1.

That is, m2=1m1.

m2=1(3)=13

Since slope of the tangent line is dydx=11+2x.

11+2x=133=1+2x

Take square on both sides,

9=1+2x91=2x8=2xx=4

Substitute x=4 in the curve y=11+2x as follows,

y=1+2(4)=1+8=9=3

Thus, the required point is (4,3).

Therefore, the point on which the curve is the tangent line perpendicular to the given line is (4, 3).

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