Respuesta :
Complete Question:
Find the directional derivative of g(x,y) = [tex]x^2y^5[/tex]at the point (1, 3) in the direction toward the point (3, 1)
Answer:
Directional derivative at point (1,3), [tex]D_ug(1,3) = \frac{162}{\sqrt{8} }[/tex]
Step-by-step explanation:
Get [tex]g'_x[/tex] and [tex]g'_y[/tex] at the point (1, 3)
g(x,y) = [tex]x^2y^5[/tex]
[tex]g'_x = 2xy^5\\g'_x|(1,3)= 2*1*3^5\\g'_x|(1,3) = 486[/tex]
[tex]g'_y = 5x^2y^4\\g'_y|(1,3)= 5*1^2* 3^4\\g'_y|(1,3)= 405[/tex]
Let P = (1, 3) and Q = (3, 1)
Find the unit vector of PQ,
[tex]u = \frac{\bar{PQ}}{|\bar{PQ}|} \\\bar{PQ} = (3-1, 1-3) = (2, -2)\\{|\bar{PQ}| = \sqrt{2^2 + (-2)^2}\\[/tex]
[tex]|\bar{PQ}| = \sqrt{8}[/tex]
The unit vector is therefore:
[tex]u = \frac{(2, -2)}{\sqrt{8} } \\u_1 = \frac{2}{\sqrt{8} } \\u_2 = \frac{-2}{\sqrt{8} }[/tex]
The directional derivative of g is given by the equation:
[tex]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} }[/tex]
