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Math
CalculusSingle Variable Calculus: Early Transcendentals, Volume I8th Edition

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Chapter 3, Problem 19P
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

Solutions

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Problem 9P
Problem 10P
Problem 11P
Problem 12P
Problem 13P
Problem 14P
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Problem 17P
Problem 18P
Problem 19P
Problem 20P
Problem 21P
Problem 22P
Problem 23P
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Let T and N be the tangent and normal lines to the ellipse x2/9 + y2/4 = 1 at any point P on the ellipse in the first quadrant. Let xT and yT be the x- and y-intercepts of T and xN and yN be the intercepts of N. As P moves along the ellipse in the first quadrant (but not on the axes), what values can .:xT, yT, xN, and yN take on? First try to guess the answers just by looking at the figure. Then use calculus to solve the problem and see how good your intuition is.

Chapter 3, Problem 19P, Let T and N be the tangent and normal lines to the ellipse x2/9 + y2/4 = 1 at any point P on the

To determine

To find: The value xT,yT,xN and yN and guess the answer from figure.

Explanation

Given:

The equation of the ellipse x29+y24=1

T and N are the tangent line and normal line to the ellipse at P in first quadent.

Calculation:

Given that xT and yT are the x and y intercepts of the tangent line T and xN and yN are the x and y intercepts of the normal line N.

From the given figure, it is observed that if the point P moves to (0,2) from right then xT approaches to infinity and yT approach 2 from right and if the point P moves to (3,0) from left then xT approaches to 3 from the right and yT approaches to infinity.

That is, the guess values form figure is xT∈(3,∞) and yT∈(2,∞) also

xN∈(0,3) and yT∈(−∞,0) or (−2,0) but it is not easy to estimate for the values xN and yN.

The slope of the tangent line to the ellipse is computed as follows,

Consider the equation of the ellipse is x29+y24=1.

Differentiate implicitly with respect to x

ddx(x29+y24)=ddx(1)ddx(x29)+ddx(y24)=019(ddx(x2))+14(ddx(y2))=019(2x)+14(2y)dydx=0

Simplify further,

y2dydx=−2x9dydx=−2x9×2ydydx=−4x9y

Thus, the derivative of the equation of the ellipse is, dydx=−4x9y.

That is, the tangent line to the ellipse at the point (x0,y0) is m=−4x09y0.

The equation of the tangen line at (x0,y0) is computed as follows,

y−y0=−4x09y0(x−x0)9y0y−9y02=−4x0x+4x024x0x+9y0y=4x02+9y02

Divide the equation by 36 on both side,

x0x9+y0y4=x029+y024

Since (x0,y0) lies on the ellipse, x029+y024=1

x0x9+y0y4=1

Therefore, the tangent line to the ellipse is, x0x9+y0y4=1.

Obtain the x-intercept to the tangent line.

Substitute y=0 in x0x9+y0y4=1

x0x9+0=1x0x=9x=9x0

Thus, the x-intercept to the tangent line xT=9x0

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P-57ESect-3.11 P-58ECh-3 P-1RCCCh-3 P-2RCCCh-3 P-3RCCCh-3 P-4RCCCh-3 P-5RCCCh-3 P-6RCCCh-3 P-7RCCCh-3 P-1RQCh-3 P-2RQCh-3 P-3RQCh-3 P-4RQCh-3 P-5RQCh-3 P-6RQCh-3 P-7RQCh-3 P-8RQCh-3 P-9RQCh-3 P-10RQCh-3 P-11RQCh-3 P-12RQCh-3 P-13RQCh-3 P-14RQCh-3 P-15RQCh-3 P-1RECh-3 P-2RECh-3 P-3RECh-3 P-4RECh-3 P-5RECh-3 P-6RECh-3 P-7RECh-3 P-8RECh-3 P-9RECh-3 P-10RECh-3 P-11RECh-3 P-12RECh-3 P-13RECh-3 P-14RECh-3 P-15RECh-3 P-16RECh-3 P-17RECh-3 P-18RECh-3 P-19RECh-3 P-20RECh-3 P-21RECh-3 P-22RECh-3 P-23RECh-3 P-24RECh-3 P-25RECh-3 P-26RECh-3 P-27RECh-3 P-28RECh-3 P-29RECh-3 P-30RECh-3 P-31RECh-3 P-32RECh-3 P-33RECh-3 P-34RECh-3 P-35RECh-3 P-36RECh-3 P-37RECh-3 P-38RECh-3 P-39RECh-3 P-40RECh-3 P-41RECh-3 P-42RECh-3 P-43RECh-3 P-44RECh-3 P-45RECh-3 P-46RECh-3 P-47RECh-3 P-48RECh-3 P-49RECh-3 P-50RECh-3 P-51RECh-3 P-52RECh-3 P-53RECh-3 P-54RECh-3 P-55RECh-3 P-56RECh-3 P-57RECh-3 P-58RECh-3 P-59RECh-3 P-60RECh-3 P-61RECh-3 P-62RECh-3 P-63RECh-3 P-64RECh-3 P-65RECh-3 P-66RECh-3 P-67RECh-3 P-68RECh-3 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