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3 years ago

Helpppppp meeee plzzzz

Mathematics

1 answer:

svlad2 [7]3 years ago

3 0

Answer:

its rather B or C

Step-by-step explanation:

c(c)=y

so the slope is Y=2x+2

the reason why its rather B or C is because of the axis

i dont know what belongs where

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Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

Parameterize the top part of the circle $x^2+y^2=1$ by

$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

with $0\le t\le\pi$ , and the line segment by

$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

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Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.

This requires

$\dfrac{\partial f}{\partial x}=\sin x\implies f(x,y)=-\cos x+g(y)$

$\dfrac{\partial f}{\partial y}=\cos y=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\sin y+C$

So we have

$f(x,y)=-\cos x+\sin y+C$

which means $\vec F$ is indeed conservative. By the fundamental theorem, we have

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=f(2,-\pi)-f(1,0)=-\cos2-(-\cos1)=\boxed{\cos1-\cos2}$

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