This is my first question on the Mathematics section of StackExchange, so please forgive me if I don't follow all the rules or things like that.
Here's my question: Consider the following potential function and graph of its equipotential curves: $\phi(x,y)=x^2+2y^2$
- Find the associated gradient field $F=\nabla\phi$
- Show that the vector field is orthogonal to the equipotential curve at the point $(1,1)$. Illustrate this result on the picture.
- Show that the vector field is orthogonal to the equipotential curve at all points $(x,y)$.
- Sketch two flow curves representing $F$ that are everywhere orthogonal to the equipotential curves.
Here's some of the equipotential curves:
Thank you so much in advance for helping me!
$\endgroup$ 12 Answers
$\begingroup$More generally, you have determined that the equipotential curves are ellipses with equation $x^2+ 2y^2= c$ for any constant, c. At any point $2x+ 4y\frac{dy}{dx}= 0$ so that $\frac{dy}{dx}= -\frac{x}{2y}$. A tangent line, at $(x_0, y_0)$, has equation $y= -\frac{x_0}{2y_0}(x- x_0)+ y_0$ so that a tangent vector is of the form $(x- x_0)\vec{i}+ (y- y_0)\vec{j}= (x- x_0)\vec{i}+ (-\frac{x_0}{2y_0}(x- x_0)\vec{j}= (\vec{i}-\frac{x_0}{2y_0}\vec{j})(x- x_0)$. And, as you know, the gradient at $(x_0, y_0)$ is $2x_0\vec{i}+ 4y_0\vec{j}$.
The dot product of $\vec{i}-\frac{x_0}{2y_0}\vec{j}$ and $2x_0\vec{i}+ 4y_0\vec{j}$ is $(1)(2x_0)-\left(\frac{x_0}{2y_0}\right)(4y_0)= 2x_0- 2x_0= 0$ so they are perpendicular.
$\endgroup$ $\begingroup$Well...
The definition of the gradient is $\nabla f\; |_{(x,y)=(a,b)}= (\frac{\partial f }{ \partial x }(a,b),\frac{\partial f }{ \partial y }(a,b))$. I leave you to calculate this value.
As for the other parts of this question, all this has to do with the fact that each level set (aka equipotential curve) has a tangent space orthogonal to the normal vector (aka the gradient). There are proofs abound on this.
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