Directional Derivative at an Angle from the Gradient

[Drakkith]

Homework Statement

(a) Find the directional derivative of z = x2y at (3,4) in the direction of 3π/4 with the x-axis. Give an exact answer.

(b) Find the directional derivative of z = x2y at (3,4) in the direction that makes an angle of 3π/4 with the gradient vector at (3,4). Give an exact answer.

(c) In which direction is the directional derivative the largest? Give your answer as a vector.

Homework Equations

$F\vec u (a,b) = \nabla F(a,b) \cdot \vec u$

Found the following two equations using the one above:
$\nabla F(3,4) = \langle 24,9 \rangle$
$\frac { \vec \nabla F(3,4)} {\left\|\vec \nabla F(3,4)\right\|} = \langle \frac {8}{\sqrt{73}},\frac {3} {\sqrt {73}} \rangle$

The Attempt at a Solution

A. I found the directional derivative as $\frac {-15}{\sqrt{2}}$.

B. I'm stuck here. I can't seem to figure out how to get the unit vector I need to dot the gradient with.
I've tried using the geometric and algebraic definitions of the dot product and the fact that$\sqrt{{u_1}^2 + {u_2}^2} = 1$ to find each component of u, but I'm just getting garbage as an answer (1 = not 1 ). Part of the problem is that I'm supposed to give an exact answer, which I haven't quite figured out how to do yet.

I just tried to use some trig logic to get the angles I need, but it doesn't seem to be working either.
Since $θ = arccos (x)$, and $θ = arcsin (y)$, I set my unit vector to $\vec u = \langle cos(\frac {3π}{4} + arccos\frac {8}{\sqrt{73}}), sin(\frac {3π}{4} + arcsin\frac {3}{\sqrt{73}})\rangle$ and dotted it with the gradient, but that's wrong as well.

C. Haven't gotten to this part yet.

The good news is that after 6 years on PF, I've finally learned the basics of LaTeX!

 

[LCKurtz]

Homework Statement

(a) Find the directional derivative of z = x2y at (3,4) in the direction of 3π/4 with the x-axis. Give an exact answer.

Do you mean $x^2y$?

(b) Find the directional derivative of z = x2y at (3,4) in the direction that makes an angle of 3π/4 with the gradient vector at (3,4). Give an exact answer.

(c) In which direction is the directional derivative the largest? Give your answer as a vector.

Homework Equations

$F\vec u (a,b) = \nabla F(a,b) \cdot \vec u$

Found the following two equations using the one above:
$\nabla F(3,4) = \langle 24,9 \rangle$
$\frac { \vec \nabla F(3,4)} {\left\|\vec \nabla F(3,4)\right\|} = \langle \frac {8}{\sqrt{73}},\frac {3} {\sqrt {73}} \rangle$

The Attempt at a Solution

A. I found the directional derivative as $\frac {-15}{\sqrt{2}}$.

B. I'm stuck here. I can't seem to figure out how to get the unit vector I need to dot the gradient with.
I've tried using the geometric and algebraic definitions of the dot product and the fact that$\sqrt{{u_1}^2 + {u_2}^2} = 1$ to find each component of u, but I'm just getting garbage as an answer (1 = not 1 ). Part of the problem is that I'm supposed to give an exact answer, which I haven't quite figured out how to do yet.

I just tried to use some trig logic to get the angles I need, but it doesn't seem to be working either.
Since $θ = arccos (x)$, and $θ = arcsin (y)$, I set my unit vector to $\vec u = \langle cos(\frac {3π}{4} + arccos\frac {8}{\sqrt{73}}), sin(\frac {3π}{4} + arcsin\frac {3}{\sqrt{73}})\rangle$ and dotted it with the gradient, but that's wrong as well.

C. Haven't gotten to this part yet.

The good news is that after 6 years on PF, I've finally learned the basics of LaTeX!

I will give you a hint on how to find that vector for (b). Let's start with $\vec v = \langle 8,3\rangle$ which is in the direction of your gradient. Two vectors perpendicular to $\vec v$ are $\langle -3,8\rangle$ and $\langle 3,-8\rangle$ (obvious from the dot product = 0). Draw these three vectors and also $-\vec v$ at the origin. Do you see in that picture two vectors you could add to give you a vector in the direction $\frac{3\pi}{4}$ from $\vec v$? If so, make it a unit vector and proceed.

 

[Drakkith]

Do you mean x^2y?

Indeed.

I will give you a hint on how to find that vector for (b). Let's start with ⃗v=⟨8,3⟩v→=⟨8,3⟩\vec v = \langle 8,3\rangle which is in the direction of your gradient. Two vectors perpendicular to ⃗vv→\vec v are ⟨−3,8⟩⟨−3,8⟩\langle -3,8\rangle and ⟨3,−8⟩⟨3,−8⟩\langle 3,-8\rangle (obvious from the dot product = 0). Draw these three vectors and also −⃗v−v→-\vec v at the origin. Do you see in that picture two vectors you could add to give you a vector in the direction 3π43π4\frac{3\pi}{4} from ⃗vv→\vec v? If so, make it a unit vector and proceed.

That worked perfectly.

B. The answer is $-\frac{657}{\sqrt{1314}}$

C. The answer is $\left\langle 24,9\right\rangle$

Thanks, Kurtz.

 

[Ray Vickson]

Indeed.
That worked perfectly.

B. The answer is $-\frac{657}{\sqrt{1314}}$

C. The answer is $\left\langle 24,9\right\rangle$

Thanks, Kurtz.

I hope you realize that for any $f(x_1, x_2, \ldots, x_n)$ the directional derivative is largest in the direction $\nabla f$, and the proof is extremely easy.

 

[Drakkith]

I hope you realize that for any $f(x_1, x_2, \ldots, x_n)$ the directional derivative is largest in the direction $\nabla f$, and the proof is extremely easy.

Yep. There's several examples in my book that I've gone through.

 

[LCKurtz]

Indeed.
That worked perfectly.

B. The answer is $-\frac{657}{\sqrt{1314}}$

Are you sure about that answer? What did you get for your unit vector in the new direction?
[Edit, added]: That answer is correct but it can be reduced.

Of course, the geometric trick I just showed you depended on the fact that the angle $\frac{3\pi}{4}$ is easy to construct geometrically. But what if you had wanted to rotate the vector clockwise by an arbitrary angle $\theta$? If you have studied the complex plane, you know that if you represent a vector $z=\langle a,b \rangle$ as $a + bi$ and multiply it by the complex number $e^{i\theta}$, the result rotates $z$ by the angle $\theta$. In your problem you would multiply by $e^{\frac {3\pi i} 4} = \cos(\frac {3\pi i} 4) + i \sin(\frac {3\pi i} 4) = -\frac 1 {\sqrt 2} + i\frac 1 {\sqrt 2}$. You could use this idea to check your work above.

 

[Drakkith]

Are you sure about that answer?

The online program accepted it as a correct answer.

What did you get for your unit vector in the new direction?

$\langle \frac {-33}{\sqrt{1314}}, \frac{15}{\sqrt{1314}}\rangle$

But what if you had wanted to rotate the vector clockwise by an arbitrary angle θθ\theta? If you have studied the complex plane, you know that if you represent a vector z=⟨a,b⟩z=⟨a,b⟩z=\langle a,b \rangle as a+bia+bia + bi and multiply it by the complex number eiθeiθe^{i\theta}, the result rotates zzz by the angle θθ\theta. In your problem you would multiply by e3πi4=cos(3πi4)+isin(3πi4)=−1√2+i1√2e3πi4=cos⁡(3πi4)+isin⁡(3πi4)=−12+i12e^{\frac {3\pi i} 4} = \cos(\frac {3\pi i} 4) + i \sin(\frac {3\pi i} 4) = -\frac 1 {\sqrt 2} + i\frac 1 {\sqrt 2}. You could use this idea to check your work above.

I have not studied the complex plane but I'll try to remember this. Thanks Kurtz.

 

[LCKurtz]

I edited my answer about the same time that you replied noting that your answer is correct but that it can be reduced. You can factor a 3 out of both numerator and denominator. Ditto for your unit vector.

 

[Drakkith]

I edited my answer about the same time that you replied noting that your answer is correct but that it can be reduced. You can factor a 3 out of both numerator and denominator. Ditto for your unit vector.

Roger!