# Find the directional derivative of f(x, y) = sqrt(xy) at P(8, 2) in the direction from P to Q(11,...

## Question:

Find the directional derivative of {eq}f(x, y) = \sqrt{xy}{/eq} at {eq} P(8, 2){/eq} in the direction from {eq}P{/eq} to {eq}Q(11, -2).{/eq}

## Gradients and Directional Derivatives:

The gradient of the function {eq}f(x, y){/eq} is given by $$ \bigtriangledown f(x,y)=f_x(x,y)\mathbf{i}+f_y(x,y)\mathbf{j}$$

To find the directional derivative of {eq}f(x, y){/eq} at a point {eq} P{/eq} in the direction of a vector, two things are necessary 1) the gradient evaluated at {eq}P{/eq} and 2) a unit vector in the correct direction.

## Answer

The partial derivative with respect to {eq}x{/eq} is {eq}f_x = y\sqrt{xy}{/eq}. The partial derivative with respect to {eq}y{/eq} is {eq}f_y = x\sqrt{xy}{/eq}. $$ \bigtriangledown f(x,y)= y\sqrt{xy} \: \mathbf{i}+ x\sqrt{xy} \: \mathbf{j}$$First evaluate the gradient at {eq} P{/eq}.

$$ \bigtriangledown f(8,2)= 2\sqrt{16} \: \mathbf{i}+ 8\sqrt{16} = 8 \: \mathbf{i}+ 32 \: \mathbf{j}$$

Find a vector in the direction from {eq}P{/eq} to {eq}Q(11, -2){/eq}. $$\overrightarrow{PQ} = \langle 11-8, -2-2\rangle = \langle 3, -4 \rangle$$

Now normalize {eq} \overrightarrow{PQ}{/eq}.$$ \| \overrightarrow{PQ} \| = \sqrt{3^2+(-4)^2} = \sqrt{25}=5 $$

A unit vector in the direction from {eq}P{/eq} to {eq}Q(11, -2){/eq} is {eq} \langle 3/5, -4/5 \rangle {/eq}. The directional derivative is $$\bigtriangledown f(8,2) \cdot \langle 3/5, -4/5 \rangle = \langle 8,32 \rangle \cdot \langle 3/5, -4/5 \rangle = 24/5 -128/5 = -104/5.$$

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