Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.4 Differentiability and Tangent Planes - Exercises - Page 789: 3


$$z=-14+5 x+10 y$$

Work Step by Step

Given $$f(x, y)= x^{2}y+xy^3 \quad \text { at } \quad(2,1)$$ Since \begin{aligned} f_{x}(x, y) &=2xy+y^3 \\ f_{x}(2,1) &= 5 \end{aligned} and \begin{aligned} f_{y}(x, y) &=x^2+3xy^2 \\ f_{y}(2,1) &= 10 \end{aligned} Then the tangent plane given by \begin{aligned} z&=f(2,1)+f_{x}(2,1)(x-2)+f_{y}(2,1)(y-1) \\ &=6+5(x-2)+10(y-1) \\ &=-14+5 x+10 y \end{aligned}
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