Directional Derivatives using Polar Coordinates

Question

I am having a hard time with this problem on my homework assignment. Here is the problem, and i will show my work below:

If $f( x, y) = -2 x^{2} + 3 y^{2}$, find the value of the directional derivative at the point $( -1, 1)$ in the direction given by the angle $\theta = \frac{2 \pi}{2}$. More specifically, find the directional derivative of f at the point $\left(-1,1\right)$ in the direction of the unit vector determined by the angle $\theta$ in polar coordinates.

To find the directional derivative, i know you need to find the partial derivative with respect to x, and then to y as shown below: $$fx(x,y)=-4x$$ $$fy(x,y)=6y$$ Then because we want the directional derivative at point P, we evaluate at point P: $$\nabla f=4i+6j$$ Now the next step is to multiply the gradient by the directional vector, and that will give the final directional derivative. I just don't know what the hint is implying when is says use the polar coordinates to find the direction of the unit vector. What i do know about polar coordinates is: $$x=rcos \theta$$ $$y=rsin\theta$$ But after that i don't know how to use it. Any help? Thanks!

Answer 1

The unit vector determined by the angle $\theta$ in polar coordinates is: $$\cos\theta\; i + \sin\theta\; j = \cos\pi\; i + \sin\pi\; j = -1\;i + 0\;j$$ I think you must have the projection of $\nabla f=4i+6j$ on this unit vector: $$(4i+6j)\cdot(-1i+0j)\times(-1i+0j) = 4i+0j$$ Here $(\cdot)$ is the inner product and $(\times)$ the scalar times vector product.

Answer 2

Nota Bene: In order to avoid possible confusion among different uses of the symbol $\theta$, I am going to let $\alpha$ denote the angle, in the polar coordinate system, of the unit vector $\hat{\mathbf n}_p$, in which direction we seek to differentiate $f(x, y)$, thus: $\hat{\mathbf n}_p = (\cos \alpha, \sin \alpha)$ at the point $p = (-1, 1)$.

Before I proceed further I think, in the interests of clarity, that I should point out that the accepted answer given by Han de Bruijn is not, in a technical sense, correct, since the directional derivative of a function with respect to a given vector is another function, not a vector as he has presented it, though it is true that $\nabla f$ can be projected along any unit vector and indeed expressed, via such projections, in any orthonormal frame. We shall in fact make use of such expression in what follows.

These things being said, I have attempted in the following to work in polar coordinates as much as possible:

The first thing to do is express the function $f(x, y)$ in polar coordinates; that's easy, using $x = r \cos \theta$ and $y = r\sin \theta$ we obtain

$f(r, \theta) = -2r^2 \cos^2 \theta + 3r^2 \sin^2 \theta = r^2(3\sin^2 \theta - 2\cos^2 \theta). \tag{1}$

Now, to get one's hands on directional derivatives in polar, or any non-Cartesian or curvilinear coordinate system, one needs to first realize that the coordinate vector fields, that is, the fields of tangent vectors to the coordinate lines or curves, are not constant, as they are in the Cartesian case, but vary from point to point, typically both in magnitude and direction. For example, in the present case the $r, \theta$ coordinate vector fields are the tangent vectors to the curves $\theta = \text{constant}$ and $r = \text{constant}$; the first are just radial lines parametrized by $r$, the second are the circles $r = \text{constant}$ parametrized by $\theta$. It should be noted that the magnitude of the radial tangent vector $\hat{\mathbf e}_r$ is $1$ everywhere, but that of the tangential field $\hat{\mathbf e}_\theta$ is in fact $r$ at the point $(r, \theta)$; this may be seen via the simple observation that an angular increment of size $\Delta \theta$ subtends an arc of length $r\Delta \theta$ on the circle of radius $r$; thus the magnitide of the tangent vector to the curve $r = \text{constant}$ is $r$ when $\theta$ is the running parameter along the curve (circle). So an orthnormal frame consistent with polar coordinates at the point $(r, \theta)$ is $\hat{\mathbf e}_r, r^{-1}\hat{\mathbf e}_\theta$. I add these remarks to try and both illustrate and drive home the point that, since the basis of coordinate vector fields varies from point to point, it will in general possess no global invariance properties as do the well-known $\mathbf i, \mathbf j, \mathbf k$ fields familiar from Cartesian coordinate systems, and the implication of this is that, in order to find the directional derivative $\mathbf v_p[g]$ of some function $g$ at some point $p$ in an arbitrary coordinate system $u_1, u_2, . . . , u_n$ (assuming for the moment we are working in $R^n$), we must first expand the direction vector $\mathbf v_p$ in terms of the basis vectors tangent to the $u_i$ curves at $p$. Once this is done, the general formula $\mathbf v_p[g] = \sum_1^n \mathbf v_p^i (\partial g / \partial u_i)$, where the $\mathbf v_p^i$ are the components of $\mathbf v_p$ in the $u_i$ coordinate basis, may be readily applied. So in the present case, what do we have? Well, (1) already presents $f$ in the polar coordinate system, so we need to work out the unit direction vector determined by the angle $\alpha$ at the point $p = (-1, 1)$, which in polar coordinates is $r_p = \sqrt 2, \theta_p = 3\pi / 4 = -5\pi / 4$. This means we need to expand the unit vector $\hat{\mathbf n}_p = (\cos \alpha, \sin \alpha)$ in terms of the unit basis vectors $\hat{\mathbf e}_r$, $r^{-1}\hat{\mathbf e}_\theta$ at the point whose Cartesian coordinates are $(-1, 1)$ and whose polars are $(\sqrt 2, 3\pi/4)$. But this is easy to do, since $\hat{\mathbf e}_r = (-1 / \sqrt{2}, 1/\sqrt{2})$ at that point and $r^{-1}\hat{\mathbf e}_\theta$ being orthonormal to $\hat{\mathbf e}_r$, may be taken as $(-1 / \sqrt{2}, -1/\sqrt{2})$ there. We have

$\hat{\mathbf n}_p = \langle \hat{\mathbf e}_r, \hat{\mathbf n}_p \rangle \hat{\mathbf e}_r + \langle r^{-1}\hat{\mathbf e}_\theta, \hat{\mathbf n}_p \rangle r^{-1}\hat{\mathbf e}_\theta, \tag{2}$

and if we now evaluate the inner products on the right-hand side of (2) we find that

$\langle \hat{\mathbf e}_r, \hat{\mathbf n}_p \rangle = \dfrac{1}{\sqrt 2}(\sin \alpha - \cos \alpha) \tag{3}$

and

$\langle r^{-1} \hat{\mathbf e}_\theta, \hat{\mathbf n}_p \rangle = -\dfrac{1}{\sqrt 2}(\cos \alpha + \sin \alpha), \tag{4}$

so that in fact

$\hat{\mathbf n}_p = \dfrac{1}{\sqrt 2}((\sin \alpha - \cos \alpha)\hat{\mathbf e}_r - (\cos \alpha + \sin \alpha)r^{-1}\hat{\mathbf e}_\theta). \tag{5}$

It should be observed that, in performing these calculations, we have employed the fact that the inner product of two vectors is an invariant independent of the frame used to evaluate it; thus the computations (3) and (4) have actually been done in a Cartesian frame, though the results are valid in the $\hat{\mathbf e}r$, $r^{-1}\hat{\mathbf e}_\theta$ basis. Using (5), it is easy to apply $\hat{\mathbf n}_p$ to $f(r, \theta)$, since

$\hat{\mathbf e}_r[f] = \dfrac{\partial}{\partial r}[f] = 2r(3\sin^2 \theta - 2\cos^2 \theta) \tag{6}$

and

$r^{-1}\hat{\mathbf e}_\theta[f] = r^{-1}\dfrac{\partial}{\partial \theta}[f] = r(6\sin \theta \cos \theta + 4\cos \theta \sin \theta) = 10r\sin \theta \cos \theta; \tag{7}$

we evaluate (6) and (7) at the point $p$, i.e. at $r_p = \sqrt 2$, $\theta_p = 3\pi / 4$, so that $\cos \theta_p = -1 / \sqrt 2$ and $\sin \theta_p = 1 / \sqrt 2$, obtaining

$\hat{\mathbf e}_r[f] = \sqrt 2 \tag{8}$

and

$r^{-1}\hat{\mathbf e}_\theta[f] = -5\sqrt 2. \tag{9}$

Now taking $\alpha = 2 \pi / 2 = \pi$ so that $\cos \alpha = -1$, $\sin \alpha = 0$ we see that

$\hat{\mathbf n}_p[f] = \dfrac{1}{\sqrt 2}(\sqrt 2 - 5\sqrt 2) = -4, \tag{10}$

the sought-for directional derivative.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!