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KengaRu [80]
3 years ago
12

Which is a nonrigid transformation of f(x)?

Mathematics
1 answer:
lianna [129]3 years ago
6 0
Nonrigid transformation changes the size but not the shape.
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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 <em>f(x, y)</em> for which

grad <em>f(x, y)</em> = <em>F</em><em>(x, y)</em>

This amounts to solving for <em>f</em> such that

∂<em>f</em>/d<em>x</em> = <em>x</em> + 3

∂<em>f</em>/∂<em>y</em> = 6<em>y</em> + 3

Integrating both sides of the first equation with respect to <em>x</em> gives

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

Differentiating with respect to <em>y</em> gives

∂<em>f</em>/∂<em>y</em> = d<em>g</em>/d<em>y</em> = 6<em>y</em> + 3

Solving for <em>g</em> gives

<em>g</em> = ∫ (6<em>y</em> + 3) d<em>y</em> = 3<em>y</em> ^2 + 3<em>y</em> + <em>C</em>

and hence

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

<em />

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

∫ <em>F</em><em>(x, y)</em> • d<em>r</em> = <em>f</em> (3, 3) - <em>f</em> (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 <em>P</em> :

∫ <em>F</em><em>(x, y)</em> • d<em>r</em> = <em>f</em> (1, 0) - <em>f</em> (0, 0) = 7/2 - 0 = 7/2

• From <em>Q</em> back to the origin:

∫ <em>F</em><em>(x, y)</em> • d<em>r</em> = <em>f</em> (0, 0) - <em>f</em> (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
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What is the SLOPE of a line PERPENDICULAR to the Given equation<br> 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}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

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

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

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

3 0
3 years ago
Which graph represents 2(cos(2 pi/3)+isin(2 pi/3))?
Artyom0805 [142]

Answer:The one with -1. B

It was D on mine tho so check the graph

Step-by-step explanation:

I got it right on edg

7 0
3 years ago
Read 2 more answers
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