Problem

10. Calculate the tangent planes of the following functions at the indicated points
(a) \( f(x, y)=x^{2}+y^{2}, p=(1,1) \)
(b) \( f(x, y)=\ln x-y^{2}, p=(e, 1) \)
(c) \( f(x, y)=x+y, p=(0,0) \)

Answer

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Answer

\[ z - f(0,0) = 1(x-0) + 1(y-0) \Rightarrow z = x + y \]

Steps

Step 1 :\[ \frac{\partial f}{\partial x} = 2x, \frac{\partial f}{\partial y} = 2y \Rightarrow \frac{\partial f}{\partial x}(1,1) = 2, \frac{\partial f}{\partial y}(1,1) = 2 \]

Step 2 :\[ z - f(1,1) = 2(x-1) + 2(y-1) \Rightarrow z - 2 = 2(x-1) + 2(y-1) \]

Step 3 :\[ \frac{\partial f}{\partial x} = \frac{1}{x}, \frac{\partial f}{\partial y} = -2y \Rightarrow \frac{\partial f}{\partial x}(e,1) = 1, \frac{\partial f}{\partial y}(e,1) = -2 \]

Step 4 :\[ z - f(e,1) = 1(x-e) - 2(y-1) \Rightarrow z - 1 = (x-e) - 2(y-1) \]

Step 5 :\[ \frac{\partial f}{\partial x} = 1, \frac{\partial f}{\partial y} = 1 \Rightarrow \frac{\partial f}{\partial x}(0,0) = 1, \frac{\partial f}{\partial y}(0,0) = 1 \]

Step 6 :\[ z - f(0,0) = 1(x-0) + 1(y-0) \Rightarrow z = x + y \]

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