Data: Atom Element Protons Neutrons Electrons Net Charge Mass Number 1 2 3 4

(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rat

vitfil [10]

Question:

Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.

(a) (i) Find the gradient of f.

(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?

(b) (i) Find the gradient of F.

(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.

Answer:

The answers to the question are

(a) (i) the gradient of f = ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.

The rate is f decreasing is -3 .

(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is ñ∙∇F = 4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)

4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .

Explanation:

f(x, y) = x²·y·eˣ⁻¹+2·x·y²

The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+ ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+ (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j

= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) at the point (1, -1) we have

∇f(x, y) = -1·i -3·j that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.

The rate is f decreasing is -3

(b) F(x, y, z) = x² + 3·y·z + 4·x·y.

The gradient of F is given by grad F(x, y, z) = ∇F(x, y, z) = = ∂f/∂x i+ ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k

The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore

ñ = ⅟4(2·i +3·j -√3·k)

The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)

= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549

8 0

3 years ago

After observing a moth that is camouflaged against dark-colored bark, a scientist asks a question. The scientist discovers that

Rasek [7]

Because of how it's worded the answer would most likely be number four

6 0

4 years ago

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Copernicus and other astronomers before him thought that celestial bodies followed a _____ orbital path.

murzikaleks [220]

The correct answer is circular. Copernicus and other astronomers before him thought that celestial bodies followed a circular orbital path. Copernicus was a Polish astronomer that concluded that the sun is at rest near the center of the universe and the earth is revolving around it annually. This theory is called heliocentric.

6 0

3 years ago

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2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$

$B_{max}=2.901\times10^{-13}\ T$

Hence, The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

7 0

3 years ago

A device consisting of four heavy balls connected by low-mass rods is free to rotate about an axle. It is initially not spinning

zubka84 [21]

The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

Using the conservation of angular momentum

$L_{i}=L_{f}$

Here, = the system's angular momentum before the collision

$L_{i}$ = 0 + mv

= (0.005)(450)(0.752)

= 1.692 kgm²/s

The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²

= 1.6292 kgm²

Here, = Iω

So,

1.692 = 1.6292(ω)

ω = 1.03 rad/s

To know more about the conservation of angular momentum, visit:

brainly.com/question/1597483

#SPJ1

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1 year ago

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