Which is a nonrigid transformation of f(x)?

Suppose F⃗ (x,y)=(x+3)i⃗ +(6y+3)j⃗ . Use the fundamental theorem of line integrals to calculate the following.

Scorpion4ik [409]

In order to use the fundamental theorem of line integrals, you need to find a scalar potential function - that is, a scalar function f(x, y) for which

grad f(x, y) = F(x, y)

This amounts to solving for f such that

∂f/dx = x + 3

∂f/∂y = 6y + 3

Integrating both sides of the first equation with respect to x gives

f = 1/2 x ^2 + 3x + g(y)

Differentiating with respect to y gives

∂f/∂y = dg/dy = 6y + 3

Solving for g gives

g = ∫ (6y + 3) dy = 3y ^2 + 3y + C

and hence

f(x, y) = 1/2 x ^2 + 3x + 3y ^2 + 3y + C

(a) By the fundamental theorem, the integral of F along any path starting at the point P (1, 0) and ending at Q (3, 3) is

∫ F(x, y) • dr = f (3, 3) - f (1, 0) = 99/2 - 7/2 = 46

(b) Now we're talking about a closed path, so the integral is simply 0. We can verify this by checking the integral over the origin-containing paths:

• From the origin to P :

∫ F(x, y) • dr = f (1, 0) - f (0, 0) = 7/2 - 0 = 7/2

• From Q back to the origin:

∫ F(x, y) • dr = f (0, 0) - f (3, 3) = 0 - 99/2 = -99/2

Then the total integral is 7/2 + 46 - 99/2 = 0, as expected.

6 0

3 years ago

katoni bought 1 1/2 dozen pencils . There are 12 pencils in a dozen . How many pencils did katoni buy?

Black_prince [1.1K]

Yessirrrrrrrrrrrrrrrr

3 0

2 years ago

Read 2 more answers

What is the SLOPE of a line PERPENDICULAR to the Given equation y = 1/3x + 1

elixir [45]

Answer:

slope for perpendicular line is -3

Step-by-step explanation:

just flip the reciprocal for the slope and change the positive/negative sign

5 0

3 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$