20 POINTS! Please solve and show work! Any problem you solve helps! Thanks!

Mathematics

1 answer:

dimulka [17.4K]4 years ago

5 0

Answer:

3. X = 13.42

Step-by-step explanation:

12² + x² = 18²

144 + x² = 324

-144 -144

x² = 180

√x² = √180

x = 13.41640 (unsure of how teacher would want you to round this, probably in the hundredths)

x = 13.42

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$\text{Show that a differentiable function f decreases most rapidly at x in the }$

$\text{direction opposite the gradient vector, that is, in the direction of}$ $-\bigtriangledown f(x)$ $\text{. Let}\ \theta \ \text{be the angle between} \bigtriangledown f(x) \ \text{and unit vector u. Then } D_u f = \mathbf{|\bigtriangledown f| \ cos \ \theta }}$

$\text{Since the minimum value of} \ \ \mathbf{cos \ \theta} \ \ is \mathbf{-1} \ \text{occuring \ for \ 0} \le \ \theta \ < 2x, \\ \\ when \ \theta = \mathbf{\pi} , \text{the mnimum value of} \ D_uf \ is} -|\bigtriangledown f|, \text{occuring when the direction of u is } \\ \\ \ \mathbf{the \ opposite \ of} \ \text{the direction of } \ \bigtriangledown f (assuming \ \bigtriangledown f\ is \ not \ zero)$

b) $\text{From part A:}$

$If \ f(x,y) = x^4y -x^2y^2 \ \ decreases \ fastest \ at \ the \point \ (2,-5)\\ \\ F(x,y) = x^4y -x^2y^3 \\ \\ f_x = \dfrac{df}{dx}= \dfrac{d}{dx}(x^4y-x^2y^3) \\ \\ f_x = \dfrac{df}{dx}= y4x^3 -2y^3x \\ \\ For(2,-5) \\ \\ f_x = (-5)4(2)^3 -2(-5)^3(2) \\ \\ \mathbf{ f_x = 340}$

$However; f_y = \dfrac{df}{dy} = \dfrac{d}{dy}(x^4y - x^2y^3) \\ \\ f_y = x^4 -3x^2y^2 \\ \\ Now, for (2, -5)\\ \\f_y = (2)^4 -3(2)^2(-5)^2 \\ \\ f_y = -284$