Find the directional derivative of the function at P in the direction of v. f(x, y) =
x/y P(1,1) v = −j
Let's find the partial derivatives:
∂f∂x=∂∂x(xy)=1y⇒∂f∂x(P)=1\frac{{\partial f}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {\frac{x}{y}} \right) = \frac{1}{y} \Rightarrow \frac{{\partial f}}{{\partial x}}\left( P \right) = 1∂x∂f=∂x∂(yx)=y1⇒∂x∂f(P)=1
∂f∂y=∂∂y(xy)=−xy2⇒∂f∂y(P)=−112=−1\frac{{\partial f}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {\frac{x}{y}} \right) = - \frac{x}{{{y^2}}} \Rightarrow \frac{{\partial f}}{{\partial y}}\left( P \right) = - \frac{1}{{{1^2}}} = - 1∂y∂f=∂y∂(yx)=−y2x⇒∂y∂f(P)=−121=−1
find the direction cosines:
∣v‾∣=02+(−1)2=1\left| {\overline v } \right| = \sqrt {{0^2} + {{( - 1)}^2}} = 1∣v∣=02+(−1)2=1
cosα=01=0, cosβ=−11=−1\cos \alpha = \frac{0}{1} = 0,\,\,\cos \beta = - \frac{1}{1} = - 1cosα=10=0,cosβ=−11=−1
Then
dfdv=∂f∂x(P)cosα+∂f∂y(P)cosβ=1⋅0−1⋅(−1)=1\frac{{df}}{{dv}} = \frac{{\partial f}}{{\partial x}}\left( P \right)\cos \alpha + \frac{{\partial f}}{{\partial y}}\left( P \right)\cos \beta = 1 \cdot 0 - 1 \cdot \left( { - 1} \right) = 1dvdf=∂x∂f(P)cosα+∂y∂f(P)cosβ=1⋅0−1⋅(−1)=1
Answer: 1
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