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 3.5, Problem 29E

(a)

To determine

To find: The equation of the tangent line to the given equation at the point.

Expert Solution

Answer to Problem 29E

The equation of the tangent line to the equation y2=5x4x2 at the point (1,2) is y=92x52.

Explanation of Solution

Given:

The curve with equation y2=5x4x2.

The point is (1,2).

Derivative rules:

(1) Chain rule: If y=f(u) and u=g(x) are both differentiable functions, then dydx=dydududx.

(2) Product rule: If y=f(u) and u=g(x) are both differentiable function, then ddx(f(x)+g(x))=ddx(f(x))+ddx(g(x)).

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (1)

where, m is the slope of the tangent line at (x1,y1) and m=dydx|x=x1,y=y1.

Calculation:

Obtain the equation of tangent line to the given point.

y2=5x4x2

Differentiate the above equation implicitly with respect to x,

ddx(y2)=ddx(5x4x2)ddx(y2)=ddx(5x4)ddx(x2)ddx(y2)=5ddx(x4)ddx(x2)ddx(y2)=20x32x

Apply the chain rule (1) and simplify the terms,

ddy(y2)dydx=20x32x2ydydx=20x32xdydx=20x32x2ydydx=10x3xy

Therefore, the derivative of the equation is dydx=10x3xy.

The slope of the tangent line at (1,2) is computed as follows,

m=dydx|(x1,y1)=(1,2)=10(1)312=92

Thus, the slope of the tangent line at (1,2) is m=92.

Substitute (1,2) for (x1,y1) and m=92 in equation (1),

y2=92(x1)y2=9x292y=9x292+2y=92x52

Therefore, the equation of the tangent line to the equation y2=5x4x2 at the point (1,2) is y=92x52.

(b)

To determine

To sketch: The curve and tangent line to the given point.

Expert Solution

Explanation of Solution

Graph:

Use online graphic calculator to draw the curve and the tangent line as shown below in Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.5, Problem 29E

From Figure 1, it is observed that the line y=92x52 touch the curve y2=5x4x2 at (1,2). That is, the tangent line y=92x52 to the curve at the point (1,2).

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