(a) Explanation Given Information: The provided equation of the curve is, x 2 + 4 y 2 − 4 = 0 . Formula used: The slope of tangent to a curve y = f ( x ) at point ( x , y ) is given by the derivative of the curve at that point. The power rule of differentiation: d d x ( x n ) = n x n − 1 Calculation: Consider the provided equation of curve, x 2 + 4 y 2 − 4 = 0 Now, find the derivative d y d x from x 2 + 4 y 2 − 4 = 0 by taking the derivative term by term on both sides of the equation as, d d x ( x 2 ) + d d x ( 4 y 2 ) − d d x ( 4 ) = d d x ( 0 ) 2 x + 8 y d y d x − 0 (b) To determine To calculate: The point at which the curve defined by x 2 + 4 y 2 − 4 = 0 has vertical tangents.

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ISBN: 9780357865095

Author: HARSHBARGER

Publisher: CENGAGE L

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Chapter 11.3, Problem 48E

(a)

To determine

To calculate: The point at which the curve defined by x2+4y2−4=0 has horizontal tangents.

(b)

To determine

To calculate: The point at which the curve defined by x2+4y2−4=0 has vertical tangents.

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Chapter 11.3, Problem 47E

Chapter 11.3, Problem 49E

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Chapter 11 Solutions

MATHEMATICAL APPLICATIONS FOR THE MGT

Ch. 11.1 - 1. Ch. 11.1 - 2. If Ch. 11.1 - Prob. 3CP Ch. 11.1 - 4. Find . Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Find the derivatives of the functions in Problems...

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The point at which the curve defined by x 2 + 4 y 2 − 4 = 0 has horizontal tangents.

Solution Summary: The author calculates the point at which the curve defined by x2+4y4=0 has horizontal tangents.

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To calculate: The point at which the curve defined by x2+4y2−4=0 has horizontal tangents.

To calculate: The point at which the curve defined by x2+4y2−4=0 has vertical tangents.

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