i have just started learning multi-variable calculus , i learned that directional derivative moves fastest along the gradient . i am not able to digest it well as for the 2-D curves that i studied a curve moves fastest along the tangent and slowest along the normal but over here it moves fastest along the gradient and slowest along the tangent plane . i am not able to understand why ? thanks for your help
1 Answers
In the case of $t \mapsto (x(t),y(t))$ the tangent to the curve is $(dx/dt, dy/dt)$. The tangent points in the direction in which $t$ increases along the curve. In contrast, a level curve $f(x,y)=k$ is the set of all $(x,y)$ which solve the equation. So, you can't compare these directly. The manner in which the curve is described is different. In the case of the level curve, if you have a parametrization $(x(t),y(t))$ of the level curve $f(x,y)=k$ then that means $f(x(t),y(t))=k$ for all $t$. But, then differentiate and use chain rule to see: $$ 0=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt} = \langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y} \rangle \cdot \langle \frac{dx}{dt},\frac{dy}{dt} \rangle. $$ This shows that the tangent vector and the gradient vector at the same point are perpendicular. The tangent vector does not really say much about how $f$ is changing, however, the magnitude of the gradient gives the max-rate of change in $f$. Many different curves take the same tangent here, but the gradient is connected to the level function $f$.
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$$\mathbf{a}\cdot\mathbf{b}=\lVert\mathbf{a}\rVert\lVert\mathbf{b}\rVert\cos \theta$$
and $\cos\theta\le 1$
with equality iff $\theta=0$ we get the result.
– Adam Hughes Sep 07 '14 at 17:44