Gradients

Relationship to Total Derivative

Recall in the article about multivariable derivatives, we talked about how single-variable derivatives approximate a line, and multivariable derivatives approximate a hyperplane.

The gradient is the tranpose of the Jacobian. That's it! $$\begin{bmatrix}\frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial x} \end{bmatrix}$$

Orthogonal to the Curve

It turns out that the gradient is orthogonal to a level curve. First, what is a level curve? $$\text{Let } w = f(x,y,z) = x^2 + y^2 + z^2$$ Setting $w = 1$ is a level curve. In fact, it is the unit sphere. $$f(x,y,z) = x^2 + y^2 + z^2$$ $$g(t) = f(x,y,z) = 1$$ $$\frac{dg}{dt} = 0 = \begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\end{bmatrix}\begin{bmatrix}x'(t) \\ y'(t) \\ z'(t)\end{bmatrix}$$ $$0 = \begin{bmatrix}\frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial z}\end{bmatrix} \cdot \begin{bmatrix}x'(t) \\ y'(t) \\ z'(t)\end{bmatrix}$$ $$0 = \text{gradient} \cdot \text{tangent line}$$ So, the gradient of that curve, which is $$\begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}$$ is orthogonal to the sphere.

Direction of Max Increase

Another important property of the gradient is that it points in the direction of greatest increase. For example, consider $f(x) = x^2 + y^2 + z^2$. Each level surface is a sphere.

If you're an ant on the surface of a unit sphere, and you want to move to a larger sphere, you have an infinite number of options. You can move along the sphere, which means your radius, or $f(x)$ value, would the stay the same. If you move toward the center, the radius decreases. If you move perpendicularly away from the sphere, the radius increases most efficiently. The gradient, the perpendicular direction, goes towards the next highest level curve.

Now, let's show this mathematically $$|D_vf(p)| = |\nabla f(p) \cdot v|$$ We can use the C-S Inequality now. $$|D_vf(p)| = |\nabla f(p) \cdot v| \leq ||\nabla f(p)||||v|| = ||\nabla f(p)|| $$ This means that the maximum value of the directional derivative is when the vector points in the direction of the gradient, which is orthogonal to the curve.

Examples

What is the tangent plane to $x^2 + y^2 = z$ at $(3, 4, 25)$? Show/Hide

The gradient is $\begin{bmatrix}2x\\3x\\-1\end{bmatrix}$ = $\begin{bmatrix}6\\8\\-1\end{bmatrix}$. Now you know the plane is of the form $$6x + 8y - z = D$$ You plug in (3,4,25) to get $$6x + 8y - z = 6(3) + 8(4) - 25 = 25$$ Now, we have our answer: $$6x + 8y - z = 25$$

You have one unit sphere at the origin. How far away do you center a sphere of radius 2 so that it's orthogonal to the first sphere? Show/Hide

The first thing to note is that if the surfaces are orthogonal, then their normal vectors are orthogonal. Therefore, the gradient of the unit sphere at the origin dot the new gradient is 0. Additionally, to simplify calculations, we can only change x. We now write two equations that we have to solve. First, we find where the normals are perpendicular: $$\begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix} \cdot \begin{bmatrix}2(x- x_0)\\2y \\ 2z\end{bmatrix}$$ $$= 4x^2 - 4xx_0 + 4y^2 + 4z^2 = 0$$ $$4(x^2 + y^2 + z^2) = 4xx_0$$ $$1 = xx_0 \to x_0 = \frac{1}{x}$$ Now, we see where the two spheres intersect $$x^2 + y^2 + z^2 = 1$$ $$(x-x_0)^2 + y^2 + z^2 = 4$$ Now, we subtract the first equation from the second. $$x^2 - 2xx_0 + x_0^2 - x^2 = 3$$ We now plug in our solution for $x = \frac{1}{x_0}$. $$x_0^2 - 2\frac{1}{x_0}x_{0} = 3$$ $$x_0 = \pm\sqrt{5}$$ We now know two things: First, the radius of the other sphere is 2 (given). Second, its center's distance from the origin is $\sqrt{5}$. <\div>