One of the tough ones..
Find a directional derivative of $f=x^2+y^2$ in the direction of $\vec{a}=\hat{i}-\hat{j}$ (vector symbols should be above the letters) at the point $(1,2)$
any ideas?
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5Removed the [tag:complex-analysis] tag. Being difficult to some does not mean it is complex anaylsis. – Emily Mar 05 '13 at 00:01
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The directional derivative of $f$ at $\mathbf{x}$ in the direction of $\mathbf{v}$ is $$\lim_{t\to 0}\frac{1}{t}\bigg( f(\mathbf{x} + t\mathbf{v}) - f(\mathbf{x})\bigg).$$ Plug, chug, repeat.
Edit: Plug in $f(\mathbf{x}) = f(x,y) = x^2 + y^2$ and $\mathbf{v} = \mathbf{i}-\mathbf{j} = (1,-1)$. So, for example, $$f(\mathbf{x} + t\mathbf{v}) = f\big( (1,2) + t(1,-1) \big) = f(1 + t, 2-t).$$ Hopefully this will get the ball rolling.
Neal
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mokoto (kusanagi??), be more specific. Do you know what $f({\bf x}+t{\bf v})$ means? Do you know what it means to plug something into an expression? Here, $t{\bf v}=(t,-t)$ and ${\bf x}=(1,2)$ so $f({\bf x}+t{\bf v})=f(1+t,2-t)=(1+t)^2+(2-t)^2$. Can you continue? Do you know how to take the derivative of a function of one variable? – anon Mar 05 '13 at 00:14
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Yes it should be the unit vector along the direction of $(1, -1)$ – Siddharth Joshi Nov 25 '15 at 19:04
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Directional Derivative = $\nabla f .\hat a $ So calculate the $\nabla f$ at the given point and take the dot product of it with the unit vector along $\vec a$.
Sahil
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