TU

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

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

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

$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

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

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

3 0

3 years ago

Help please need this answer

fredd [130]

Answer:

(5,4) is the solution

Step-by-step explanation:

2x + y = 14 --> y = -2x + 14

Substitute y = -2x + 14 into 3x - 2y = 7

3x - 2( -2x + 14) = 7

3x + 4x - 28 = 7

7x = 35

x = 5

y = -2(5) + 14

y = -10 + 14

y = 4

Answer (5 , 4)

7 0

3 years ago

Read 2 more answers

Owen and Eva spent the same amount of money making copies. Owen spent 14 cents per copy, while Eva spent 10 cents per copy. What

melamori03 [73]

Answer:

2

Step-by-step explanation:

14 times 5= 70

10 times 7= 70

70 is the LCM for 14 and 10, least common multiple meaning the least difference, then subtract to get 2

i hope this helps u pls give a brainliest and a thx ;)

5 0

3 years ago

Read 2 more answers

Pls help me I dummie lol

vivado [14]

The missing number is -1

6 0

3 years ago

Read 2 more answers

A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp

abruzzese [7]

Step-by-step explanation:

"A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp."

"(a) During a morning activity of the camp, these 12 players have to randomly group into six pairs of two players each."

"(i) Find the total number of possible ways that these six pairs can be formed."

The order doesn't matter (AB is the same as BA), so use combinations.

For the first pair, there are ₁₂C₂ ways to choose 2 people from 12.

For the second pair, there are ₁₀C₂ ways to choose 2 people from 10.

So on and so forth. The total number of combinations is:

₁₂C₂ × ₁₀C₂ × ₈C₂ × ₆C₂ × ₄C₂ × ₂C₂

= 66 × 45 × 28 × 15 × 6 × 1

= 7,484,400

"(ii) Find the probability that each pair contains players of the same gender only. Correct your final answer to 4 decimal places."

We need to find the number of ways that 6 boys can be grouped into 3 pairs. Using the same logic as before:

₆C₂ × ₄C₂ × ₂C₂

= 15 × 6 × 1

= 90

There are 90 ways that 6 boys can be grouped into 3 pairs, which means there's also 90 ways that 6 girls can be grouped into 3 pairs. So the probability is:

90 × 90 / 7,484,400

= 1 / 924

≈ 0.0011

"(b) During an afternoon activity of the camp, 6 players are randomly selected and 6 one-on-one matches with the coach are to be scheduled.

(i) How many different schedules are possible?"

There are ₁₂C₆ ways that 6 players can be selected from 12. From there, each possible schedule has a different order of players, so we need to use permutations.

There are 6 options for the first match. After that, there are 5 options for the second match. Then 4 options for the third match. So on and so forth. So the number of permutations is 6!.

The total number of possible schedules is:

₁₂C₆ × 6!

= 924 × 720

= 665,280

"(ii) Find the probability that the number of selected male players is higher than that of female players given that at most 4 females were selected. Correct your final answer to 4 decimal places."

If at most 4 girls are selected, that means there's either 0, 1, 2, 3, or 4 girls.

If 0 girls are selected, the number of combinations is:

₆C₆ × ₆C₀ = 1 × 1 = 1

If 1 girl is selected, the number of combinations is:

₆C₅ × ₆C₁ = 6 × 6 = 36

If 2 girls are selected, the number of combinations is:

₆C₄ × ₆C₂ = 15 × 15 = 225

If 3 girls are selected, the number of combinations is:

₆C₃ × ₆C₃ = 20 × 20 = 400

If 4 girls are selected, the number of combinations is:

₆C₂ × ₆C₄ = 15 × 15 = 225

The probability that there are more boys than girls is:

(1 + 36 + 225) / (1 + 36 + 225 + 400 + 225)

= 262 / 887

≈ 0.2954

7 0

2 years ago