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2 years ago

5

The table shows the number of cups of flour, f, that a bakery needs for the number of pound cakes that they make, p. 3 6 9 14 Po

und Cakes, P Cups of Flour, f 8.25 16.5 24.75

1 answer:

Mumz [18]2 years ago

4 0

Ok.

3 8.25

6 16.5

9 24.75

14

8.25/3=2.75

14*2.75=

<h2><u><em>38.5</em></u></h2>

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Find the directional derivative of at the point (1, 3) in the direction toward the point (3, 1). g

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Find the directional derivative of g(x,y) = $x^2y^5$ at the point (1, 3) in the direction toward the point (3, 1)

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Directional derivative at point (1,3), $D_ug(1,3) = \frac{162}{\sqrt{8} }$

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Get $g'_x$ and $g'_y$ at the point (1, 3)

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$g'_x = 2xy^5\\g'_x|(1,3)= 2*1*3^5\\g'_x|(1,3) = 486$

$g'_y = 5x^2y^4\\g'_y|(1,3)= 5*1^2* 3^4\\g'_y|(1,3)= 405$

Let P = (1, 3) and Q = (3, 1)

Find the unit vector of PQ,

$u = \frac{\bar{PQ}}{|\bar{PQ}|} \\\bar{PQ} = (3-1, 1-3) = (2, -2)\\{|\bar{PQ}| = \sqrt{2^2 + (-2)^2}\\$

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The unit vector is therefore:

$u = \frac{(2, -2)}{\sqrt{8} } \\u_1 = \frac{2}{\sqrt{8} } \\u_2 = \frac{-2}{\sqrt{8} }$

The directional derivative of g is given by the equation:

$D_ug(1,3) = g'_x(1,3)u_1 + g'_y(1,3)u_2\\D_ug(1,3) = (486*\frac{2}{\sqrt{8} } ) + (405*\frac{-2}{\sqrt{8} } )\\D_ug(1,3) = (\frac{972}{\sqrt{8} } ) + (\frac{-810}{\sqrt{8} } )\\D_ug(1,3) = \frac{162}{\sqrt{8} }$

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