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2 years ago

-9 - (18v - 5) = 2(7v - 16) - 3v

Mathematics

1 answer:

andrezito [222]2 years ago

6 0

Answer:

v = 28/29

Step-by-step explanation:

- 9 - (18v - 5) = 2(7v - 16) - 3v

- 9 - 18v + 5 = 14v - 32 - 3v

- 18v - 14v + 3v = - 32 - 5 + 9

- 29v = - 28

v = -28 / -29

v = 28/29

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If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (2, 4, 0) in the direction

valentina_108 [34]

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a) $\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$ .

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Step-by-step explanation:

Given a function $f(x,y,z)$ , this function has the following gradient:

$\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}$ .

(a) find the gradient of f

We have that $f(x,y,z) = x\sin{yz}$ . So

$f_{x}(x,y,z) = \sin{yz}$

$f_{y}(x,y,z) = xz\cos{yz}$

$f_{z}(x,y,z) = xy \cos{yz}$ .

$\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}$ .

$\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$

(b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k.

The directional derivate is the scalar product between the gradient at (2,4,0) and the unit vector of v.

We have that:

$\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$

$\nabla f(2,4,0) = \sin{0}\mathbf{i} + 0\cos{0}\mathbf{j} + 8 \cos{0}\mathbf{k}$ .

$\nabla f(2,4,0) = 0i+0j+8k=(0,0,8)$

The vector is $v = i + 3j - k = (1,3,-1)$

To use v as an unitary vector, we divide each component of v by the norm of v.

$|v| = \sqrt{1^{2} + 3^{2} + (-1)^{2}} = \sqrt{11}$

$v_{u} = (\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}})$

Now, we can calculate the scalar product that is the directional derivative.

$Du_{f}(2,4,0) = (0,0,8).(\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}) = -\frac{8}{\sqrt{11}}$

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3 years ago

-9 - (18v - 5) = 2(7v - 16) - 3v​

-9 - (18v - 5) = 2(7v - 16) - 3v