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3 years ago

6xb is less than 12 but greater than 6

1 answer:

7 0

I interpret your "xb" to mean "6 times b." Please do not use " x " to indicate multiplication. Use " * " if you wish.

6 < 6b < 12, which can be reduced to 1 < b < 2

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Consider the following. C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

horsena [70]

Answer:

$\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}$

$\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}$

$\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}$

$\mathbf{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}$

Step-by-step explanation:

Given that:

C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

a. Find a piecewise smooth parametrization of the path C.

r(t) = { 0

If C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1),

Then:

$C_1 = (0,0) \\ \\ C_2 = (1,0) \\ \\ C_3 = (0,1)$

Also:

$\mathtt{r_1 = (0,0) + t(1,0) = (t,0) }$

$\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}$

$\mathtt{r_2 = (1,0) + t(-1,1) = (1- t,t) }$

$\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}$

$\mathtt{r_3 = (0,1) + t(0,-1) = (0,1-t) }$

$\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}$

b Evaluate :

Integral of (x+2y^1/2)ds

$\mathtt{\int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \int \limits ^1_{0} \ (t + 0) \sqrt{1} } \\ \\ \mathtt{ \int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \begin {pmatrix} \dfrac{t^2}{2} \end {pmatrix} }^1_0 \\ \\ \mathtt{\int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \dfrac{1}{2}}$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \int \limits (x+2 \sqrt{y} \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2 \ dt } }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \int \limits 2- t + 2\sqrt{t-1} \ \sqrt{1+1} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} \int \limits^2_1 2- t + 2\sqrt{t-1} \ dt }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2t - \dfrac{t^2}{2}+ \dfrac{2(t-1)^{3/2}}{3} (2) \end {pmatrix} ^2_1}$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2 -\dfrac{1}{2} (4-1)+\dfrac{4}{3} (1)^{3/2} -0 \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2 -\dfrac{3}{2} + \dfrac{4}{3} \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} \dfrac{12-9+8}{6} \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} \dfrac{11}{6} \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \dfrac{ \sqrt{2} }{6} \ (11 )}$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \dfrac{ 11 \sqrt{2} }{6}}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits ^3_2 0+2 \sqrt{3-t} \ \sqrt{0+1} }$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits ^3_2 2 \sqrt{3-t} \ dt}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits^3_2 \begin {pmatrix} \dfrac{-2(3-t)^{3/2}}{3} (2) \end {pmatrix}^3_2 }$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [(0)-(1)]}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [-(1)]}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \dfrac{4}{3}}$

$\mathtt{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}}{6}+\dfrac{1}{2}+ \dfrac{4}{3}}$

$\mathtt{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+3+8}{6}}$

$\mathbf{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}$

5 0

3 years ago

The function f(x,y)=2x + 2y has an absolute maximum value and absolute minimum value subject to the constraint 9x^2 - 9xy + 9y^

deff fn [24]

Answer:

the minimum is located in x = -5/3 , y= -5/3

Step-by-step explanation:

for the function

f(x,y)=2x + 2y

we define the function g(x)=9x² - 9xy + 9y² - 25 ( for g(x)=0 we get the constrain)

then using Lagrange multipliers f(x) is maximum when

fx-λgx(x)=0 → 2 - λ (9*2x - 9*y)=0 →

fy-λgy(x)=0 → 2 - λ (9*2y - 9*x)=0

g(x) =0 → 9x² - 9xy + 9y² - 25 = 0

subtracting the second equation to the first we get:

2 - λ (9*2y - 9*x) - (2 - λ (9*2x - 9*y))=0

- 18*y + 9*x + 18*x - 9*y = 0

27*y = 27 x → x=y

thus

9x² - 9xy + 9y² - 25 = 0

9x² - 9x² + 9x² - 25 = 0

9x² = 25

x = ±5/3

thus

y = ±5/3

for x=5/3 and y=5/3 → f(x)= 20/3 (maximum) , while for x = -5/3 , y= -5/3 → f(x)= -20/3 (minimum)

finally evaluating the function in the boundary , we know because of the symmetry of f and g with respect to x and y that the maximum and minimum are located in x=y

thus the minimum is located in x = -5/3 , y= -5/3

4 0

4 years ago

Can u help me wirh some questions please

Pavel [41]

Find the closest whole number estimate for 3.8 and 6.1

The estimate for both numbers are, 4 and 6.

3.8 is rounded up to the nearest whole number, while 6.1 is rounded down to 6because the 1 after the decimal is an insignificant number (its less than 5)

5 0

1 year ago

GaryK [48]

Answer:

5/6

Step-by-step explanation:

4/6 + 1/6 = 5/6

4 0

3 years ago

Simplify the expression 4b-24+19

Tpy6a [65]

4b - 24 + 19 = 4b - (-24 + 19) = 4b - 5

4 0

3 years ago