   Chapter 3.5, Problem 33E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# (a) The curve with equation y2 = 5x4 – x2 is called a kampyle of Eudoxus. Find an equation of the tangent line to this curve at the point (1, 2).(b) Illustrate part (a) by graphing the curve and the tangent line on a common screen. (If your graphing device will graph implicitly defined curves, then use that capability. If not, you can still graph this curve by graphing its upper and lower halves separately.)

(a)

To determine

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

Explanation

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

(b)

To determine

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

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