Explanation Given: x 2 + 2 x y − y 2 + x = 2 Concept Used: To find an equation of the tangent line to the graph of an equation y = f ( x ) at the a point ( a , b ) First find slope of the equation at the prescribed point as: m = d y d x | ( a , b ) Now, the equation of the tangent line to the graph of the equation is y = m x + c Put ( a , b ) coordinate in y = m x + c to find the value of c Calculation: To find an equation of the tangent line to the curve x 2 + 2 x y − y 2 + x = 2 at ( 1 , 2 ) First find slope of the equation at the prescribed point as: m = d y d x | ( a , b ) For that differentiate both side of the equation with respect to x as d d x ( x 2 + 2 x y − y 2 + x ) = d d x ( 2 ) d d x x 2 + 2 d d x ( x y ) − d d x y 2 + d d x x = 0 2 x + 2 x d d x y + 2 y d x d x + 2 y d y d x

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ISBN: 9781305420878

Author: Stewart

Publisher: CENGAGE L

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Chapter 3.5, Problem 78E

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To find an equation of the tangent line to the hyperbola x2+2xy−y2+x=2 at (1,2)

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Chapter 3.5, Problem 77E

Chapter 3.5, Problem 79E

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Question 2 Let F be a solenoidal vector field, suppose V × F = (-8xy + 12z², −9x² + 4y² + 9z², 6y²), and let (P,Q,R) = V²F(.725, —.283, 1.73). Then the value of sin(2P) + sin(3Q) + sin(4R) is -2.024 1.391 0.186 -0.994 -2.053 -0.647 -0.588 -1.851 1 pts

1 pts Let F and G be vector fields such that ▼ × F(0, 0, 0) = (0.76, -9.78, 3.29), G(0, 0, 0) = (−3.99, 6.15, 2.94), and G is irrotational. Then sin(5V (F × G)) at (0, 0, 0) is Question 1 -0.246 0.072 -0.934 0.478 -0.914 -0.855 0.710 0.262 .

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Chapter 3 Solutions

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To find an equation of the tangent line to the hyperbola x 2 + 2 x y − y 2 + x = 2 at ( 1 , 2 )

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To find an equation of the tangent line to the hyperbola x2+2xy−y2+x=2 at (1,2)

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