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This question confused me as well. The only thing I could think was that you could find a $\vec u$ that was perpendicular to $\langle 3,4 \rangle$ so you could do $\langle 4,-3\rangle$ or $\langle -4,3\rangle$. But when I used those I didn't come up with the same answer as the book. So hopefully someone will be able to clarify this for both of us.

This question confused me as well. The only thing I could think was that you could find a $\vec u$ that was perpendicular to $\langle 3,4 \rangle$ so you could do $\langle 4,-3\rangle$ or $\langle -4,3\rangle$. But when I used those I didn't come up with the same answer as the book. So hopefully someone will be able to clarify this for both of us.

This question confused me as well. The only thing I could think was that you could find a $\vec u$ that was perpendicular to $\langle 3,4 \rangle$ so you could do $\langle 4,-3\rangle$ or $\langle -4,3\rangle$. But when I used those I didn't come up with the same answer as the book. So hopefully someone will be able to clarify this for both of us.

Following Christina's points of (-3,-4) $$ f(x,y) = ln\sqrt{(x^2+y^2)}$$ $$ D_\vec u f = \nabla f \cdot \vec u $$ $$ \nabla f = \langle \frac{ln(x)}{\sqrt{(x^2+y^2)}}, \frac{ln(y)}{\sqrt{(x^2+y^2)}}\rangle$$ $$ \nabla f(x_0,y_0) = \langle \frac{ln(3)}{\sqrt{(3^2+4^2)}}, \frac{ln(4)}{\sqrt{(3^2+4^2)}}\rangle$$ $$ = \langle \frac {ln(3)}{5} , \frac {ln(4)}{5}\rangle$$ $$ \vec u = \langle -3,-4\rangle$$ $$ D_\vec u f = \langle \frac {ln(3)}{5} , \frac {ln(4)}{5}\rangle \cdot \langle -3,-4 \rangle$$ $$ = -\frac{3ln(3)}{5} - \frac{4ln(4)}{5}$$Thanks Christina! I need to go back and restudy how to do my derivatives apparently :(