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Find the directional derivative of $f(x,y,z)=zx+y^4$ at the point $(1,3,2)$ in the direction of a vector making an angle of $π/4$ with $\nabla f(1,3,2)$.

Alapan Das
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jackbeen
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  • What have you tried? What is the formula for a directional derivative? You will need to calculate the gradient of the function and evaluate it at $(1, 3, 2)$. – paulinho May 19 '20 at 13:45

1 Answers1

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First find the partial derivatives which gives the gradient.

$\frac{\partial{f}}{\partial{x}}=z$ ,$\frac{\partial{f}}{\partial{y}}=4y^3$ and $\frac{\partial{f}}{\partial{z}}=x$.

Hence $\triangledown f=z\hat{i}+4y^3\hat{j}+x\hat{k} \Rightarrow \triangledown f(1,3,2)=2\hat{i}+108\hat{j}+1\hat{k}$

Let's find unit vector along $\triangledown f$, $\hat{n}=108.02[2\hat{i}+4\hat{j}+\hat{k}]$.

We need to find the directional derivative along some $\hat{u}$ which is given by:

$D_u f(x,y,z)=\triangledown f(x,y,z) \cdot \hat{u} = |\triangledown f(x,y,z)| cos(\theta)$, where $\theta$ is the angle between the vectors $\hat{u}$ and $\triangledown f(x,y,z)$ which is $\frac{\pi}{4}$

Hence $D_uf(1,3,2)=108.02 \frac{1}{\sqrt{2}}=76.38$.

CNiD
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