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Let $f(x,y)=4x^3y^2$

How do I find the directional derivative of $f$ at $(2,1)$ in the direction of the vector $3i-4j$? What would be a unit vector in the direction in which $f$ decreases most rapidly from the point $(2,1)$? And, what is the rate of change of $f$ in the direction given in the second question?

My Work (What I have done so far):

For the first question: $$\nabla f(x,y)=12x^2y^2\mathbf i+8x^3y\mathbf j$$ $$\nabla f(2,1)=48\mathbf i+64\mathbf j$$ $$u=(3/5)\mathbf i-(4/5)\mathbf j$$

therefore $D_u f=\nabla f\bullet u=0$.

I'm not entirely sure if this is right? Can someone please verify?

For the second question: $$u=(3/5)\mathbf i-(4/5)\mathbf j$$

therefore $$-\|\nabla f(2,1)\|=?$$

How do I find this? And am I doing this right?

I have no idea what to do for the third question.

Mikeal
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  • For the first question, you want to compute $\nabla f(2,1) \cdot u$ where $u$ is the unit vector of $(3,-4)$. For your second question, remember that the gradient will point in the direction in which $f$ increased the most. So $-\nabla f(2,1)$ is the direction you're looking for. – MathNewbie May 22 '15 at 01:17
  • @MathNewbie I have attached my work can you look at that? – Mikeal May 22 '15 at 01:19
  • The idea looks correct. Just double check to see if you've made a mistake with your computation just in case. – MathNewbie May 22 '15 at 01:22
  • @Mikeal It looks like you've got the right idea. But for the first question: make sure you multiplied out the dot product correctly (hint: you didn't). For the second question remember it's just $-u$ because it's the direction of maximum decrease not increase. For the third question, you already calculated $\nabla f(2,1)$ so all you need to do then is to find the magnitude just like you wrote -- do you know how to do that? –  May 22 '15 at 01:32
  • @Bye_World No I think i forgot. How do you do it? – Mikeal May 22 '15 at 01:37
  • Example: If $f(x,y,z) = 3x^2\mathbf i + 2xy\mathbf j - 3\mathbf k$, then $$|f(x,y,z)| = \sqrt{(3x^2)^2 + (2xy)^2 + (-3)^2}$$ So $|f(1,0,1)| = \sqrt{18}$. –  May 22 '15 at 01:39
  • In the first question, you are asked to find the directional derivative in the given direction, not the normalised direction. – copper.hat May 22 '15 at 03:33
  • Please do not ask the (exact) same question multiple times. You should instead edit your previous question to add any progress you've made or further remarks. – A.P. May 27 '15 at 09:55

1 Answers1

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The one sided (take two sided if more appropriate, it doesn't matter here) derivative is defined as $df(x,h) = \lim_{t \downarrow 0} {f(x+th)-f(x) \over t}$.

It is not hard to show that if $f$ is differentiable at $x$, then $df(x,h) = \langle \nabla f(x), h \rangle$. Or you can compute directly from the formula above.

Aside: If $u$ is a unit vector, then $df(x,u)$ gives the rate of change of $f$ in the $u$ direction.

We have $f(x) = 4 x_1^3 x_2^2$, $x= (2,1)^T, h = (3,-4)^T$, so $df(x,h) = -112$

The direction of steepest descent is given by the negative of the gradient at the given point, that is $g=-\nabla f(x) = (-48,-64)^T$, to get a unit vector, we divide by the length $\|g\| = 80$.

To get the rate of descent in this direction, we calculate $df(x,{1 \over \|g\|}g) = \langle \nabla f(x), -{1 \over \|\nabla f(x)\| }\nabla f(x) \rangle = - \|\nabla f(x)\|$, hence we get $-80$.

copper.hat
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