Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 2.4, Problem 36E

To determine

To find: The equations of all lines tangent to y=9−x2 that pass through the point (1,12).

Expert Solution & Answer

Answer to Problem 36E

The equations of all lines tangent to y=9−x2 that pass through the point (1,12) are y=2x+10 and y=−6x+18.

Explanation of Solution

Given information:

The curve is y=9−x2 and the passing point is (1,12).

Calculation:

The formula for the slope of curve at x=a is equal to limh→0f(a+h)−f(a)h.

Substitute a+h for x in the function.

f(x)=9−x2f(a+h)=9−(a+h)2f(a+h)=9−(a2+h2+2ah)f(a+h)=9−a2−h2−2ah

Substitute a for x in the function.

f(x)=9−x2f(a)=9−a2

Substitute 9−a2−h2−2ah for f(a+h) and 9−a2 for f(a) in the formula for slope.

limh→0f(a+h)−f(a)h=limh→09−a2−h2−2ah−(9−a2)h=limh→0−h2−2ahh=limh→0h(−h−2a)h=limh→0(−h−2a)

Further simplify.

limh→0f(a+h)−f(a)h=0−2a=−2a

The tangent line to the curve y=9−x2 at the point x=a.

y−f(a)=m(x−a)y−(9−a2)=−2a(x−a)y−9+a2=−2ax+2a2y=−2ax+a2+9

As the tangent line passes through the point (1,12) . Substitute 1 for x and 12 for y in the tangent line.

y=−2ax+a2+912=−2a(1)+a2+912=−2a+a2+9a2−2a−3=0

Further simplify,

a2−2a−3=0a2−3a+a−3=0a(a−3)+1(a−3)=0(a+1)(a−3)=0

So, the values of a are −1 and 3.

Substitute −1 for a in the tangent line.

y=−2ax+a2+9y=−2(−1)x+(−1)2+9y=2x+1+9y=2x+10

Substitute 3 for a in the tangent line.

y=−2ax+a2+9y=−2(3)x+(3)2+9y=−6x+9+9y=−6x+18

Therefore, the equations of all lines tangent to y=9−x2 that pass through the point (1,12) are y=2x+10 and y=−6x+18.

Chapter 2.4, Problem 35E

Chapter 2.4, Problem 37E

Chapter 2 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 2.1 - Prob. 1QR Ch. 2.1 - Prob. 2QR Ch. 2.1 - Prob. 3QR Ch. 2.1 - Prob. 4QR Ch. 2.1 - Prob. 5QR Ch. 2.1 - Prob. 6QR Ch. 2.1 - Prob. 7QR Ch. 2.1 - Prob. 8QR Ch. 2.1 - Prob. 9QR Ch. 2.1 - Prob. 10QR

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