A shopping cart is initially at rest when it undergoes a constant acceleration. After

Find a unit vector in the direction in which f increases most rapidly at P and give the rate of chance of f in that direction; f

Burka [1]

Answer:

Check attachment for complete question

Question

Find a unit vector in the direction in which

f increases most rapidly at P and give the rate of change of f

in that direction; Find a unit vector in the direction in which f

decreases most rapidly at P and give the rate of change of f in

that direction.

f (x, y, z) = x²z e^y + xz²; P(1, ln 2, 2).

Explanation:

The function, z = f(x, y,z), increases most rapidly at (a, b,c) in the

direction of the gradient and decreases

most rapidly in the opposite direction

Given that

F=x²ze^y+xz² at P(1, In2, 2)

1. F increases most rapidly in the positive direction of ∇f

∇f= df/dx i + df/dy j +df/dz k

∇f=(2xze^y+z²)i + (x²ze^y) j + (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= (2×1×2×e^In2+2²)i +(1²×2×e^In2)j +(1²e^In2+2×1×2)

∇f=12i + 4j + 6k

Then, unit vector

V= ∇f/|∇f|

Then, |∇f|= √ 12²+4²+6²

|∇f|= 14

Then,

Unit vector

V=(12i+4j+6k)/14

V=6/7 i + 2/7 j + 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

2. F increases most rapidly in the positive direction of -∇f

∇f=- (df/dx i + df/dy j +df/dz k)

∇f=-(2xze^y+z²)i - (x²ze^y) j - (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= -(2×1×2×e^In2+2²)i -(1²×2×e^In2)j -(1²e^In2+2×1×2)

∇f=-12i -4j - 6k

Then, unit vector

V= -∇f/|∇f|

Then, |∇f|= √ 12²+4²+6²

|∇f|= 14

Then,

Unit vector

V=-(12i+4j+6k)/14

V= - 6/7 i - 2/7 j - 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

6 0

4 years ago

Read 2 more answers

What describes Newton’s law of universal gravitation

VikaD [51]

Newton's law of universal gravitation gives the gravitational force between two objects:

F = GMm/r²

F = gravitational force, G = gravitational constant, M & m are the masses of the two objects, r = distance between the objects

8 0

4 years ago

Carbon-14 is used to determine the age of ancient objects. If a sample today contains 0.060 g of carbon-14, how much carbon-14 m

MakcuM [25]

Answer: 86.47 g of carbon-14 must have been present in the sample 11,430 years ago.

Explanation:

Half-life of sample of carbon -14= 5,730 days

$\lambda=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{5,730 days}=0.00012 day^{-1}$

Let the sample present 11,430 years(t) ago = $N_o$

Sample left till today ,N= 0.060 g

$N=N_o\times e^{-\lambda t}$

$ln[N]=ln[N]_o-\lambda t$

$\log[0.060 g]=\log[N_o]-2.303\times 0.00012 day^{-1}\times 11,430 days$

$\log[N_o]=1.9369$

$N_o=86.47 g$

86.47 g of carbon-14 must have been present in the sample 11,430 years ago.

5 0

3 years ago

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The latitude of any location on earth is the angle formed by the two rays drawn from the center of earth to the location and to

Alina [70]

The distance between city a and city b is 833.345 miles.

We know that

1°=60'

The distance of city a from the initial ray is calculated as

x_a=3960*tan45.46°=4024.101 miles

The distance of city b from the initial ray is calculated as

x_b=3960*tan 38.86°=3190.75 miles

Now the distance between city a and b is equal to

4024.101-3190.75=833.345 miles

This is the vertical distance between the cities.

5 0

3 years ago

A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upwa

Ksju [112]

The toy rocket is launched vertically from ground level, at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance.

The total energy of the rocket, which is a sum of its kinetic energy and potential energy, is constant.

At a height of 72 m with the rocket moving at 30 m/s, the total energy is m*9.8*72 + (1/2)*m*30^2 where m is the mass of the rocket.

At ground level, the total energy is 0*m*9.8 + (1/2)*m*v^2.

Equating the two gives: m*9.8*72 + (1/2)*m*30^2 = 0*m*9.8 + (1/2)*m*v^2

=> 9.8*72 + (1/2)*30^2 = (1/2)*v^2

=> v^2 = 11556/5

=> v = 48.07

<span>The velocity of the rocket when it impacts the ground is 48.07 m/s</span>

3 0

3 years ago