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Paraphin [41]
3 years ago
10

Answer the question ​

Mathematics
1 answer:
guajiro [1.7K]3 years ago
6 0

what is the questionswjjbdbfnjr

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Find the directional derivative of at the point (1, 3) in the direction toward the point (3, 1). g
Anika [276]

Complete Question:

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

Answer:

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

Step-by-step explanation:

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

g(x,y) = x^2y^5

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}\\

|\bar{PQ}| = \sqrt{8}

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} }

8 0
3 years ago
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