4.6 Directional derivatives and the gradient  (Page 4/8)

[link] shows a portion of the graph of the function $f\left(x,y\right)=3+\text{sin}x\text{sin}y.$ Given a point $\left(a,b\right)$ in the domain of $f,$ the maximum value of the gradient at that point is given by $\Vert \nabla f\left(a,b\right)\Vert .$ This would equal the rate of greatest ascent if the surface represented a topographical map. If we went in the opposite direction, it would be the rate of greatest descent.

When using a topographical map, the steepest slope is always in the direction where the contour lines are closest together (see [link] ). This is analogous to the contour map of a function, assuming the level curves are obtained for equally spaced values throughout the range of that function.

Gradients and level curves

Recall that if a curve is defined parametrically by the function pair $\left(x\left(t\right),y\left(t\right)\right),$ then the vector $x^{\prime }\left(t\right)i+y^{\prime }\left(t\right)j$ is tangent to the curve for every value of $t$ in the domain. Now let’s assume $z=f\left(x,y\right)$ is a differentiable function of $x\text{and}y,$ and $\left(x_0,y_0\right)$ is in its domain. Let’s suppose further that $x_0=x\left(t_0\right)$ and $y_0=y\left(t_0\right)$ for some value of $t,$ and consider the level curve $f\left(x,y\right)=k.$ Define $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)$ and calculate $g^{\prime }\left(t\right)$ on the level curve. By the chain Rule,

But $g^{\prime }\left(t\right)=0$ because $g\left(t\right)=k$ for all $t.$ Therefore, on the one hand,

on the other hand,

Therefore,

Thus, the dot product of these vectors is equal to zero, which implies they are orthogonal. However, the second vector is tangent to the level curve, which implies the gradient must be normal to the level curve, which gives rise to the following theorem.

We can use this theorem to find tangent and normal vectors to level curves of a function.

Finding tangents to level curves

For the function $f\left(x,y\right)=2x^2-3xy+8y^2+2x-4y+4,$ find a tangent vector to the level curve at point $\left(\mathrm{-2},1\right).$ Graph the level curve corresponding to $f\left(x,y\right)=18$ and draw in $\nabla f\left(\mathrm{-2},1\right)$ and a tangent vector.

First, we must calculate $\nabla f\left(x,y\right)\text{:}$

$f_x\left(x,y\right)=4x-3y+2\text{and}{f}_y=\mathrm{-3}x+16y-4\text{so}\nabla f\left(x,y\right)=\left(4x-3y+2\right)i+\left(\mathrm{-3}x+16y-4\right)j.$

Next, we evaluate $\nabla f\left(x,y\right)$ at $\left(\mathrm{-2},1\right)\text{:}$

$\nabla f\left(\mathrm{-2},1\right)=\left(4\left(\mathrm{-2}\right)-3\left(1\right)+2\right)i+\left(\mathrm{-3}\left(\mathrm{-2}\right)+16\left(1\right)-4\right)j=\mathrm{-9}i+18j.$

This vector is orthogonal to the curve at point $\left(\mathrm{-2},1\right).$ We can obtain a tangent vector by reversing the components and multiplying either one by $\mathrm{-1}.$ Thus, for example, $\mathrm{-18}i-9j$ is a tangent vector (see the following graph).

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