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victus00 [196]
2 years ago
15

I need help with this its timed so asap.

Mathematics
2 answers:
qaws [65]2 years ago
5 0

Answer:

Height of cactus= 3.8ft

Step-by-step explanation:

Height of palm tree= 4.9ft

Height of cactus= ?

we know, 4.9ft= 1.2ft+c

   So, 4.9ft- 1.2ft= c

    c= 3.8ft

mixer [17]2 years ago
4 0
Height of cactus is 3.8 ft
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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]

Answer:

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

b) Du_{f}(2,4,0) = -\frac{8}{\sqrt{11}}

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}

So

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}}

6 0
3 years ago
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