What is the index of the radical below?

Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l

SIZIF [17.4K]

Answer:

$\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}$

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let

$C_1,C_2$

be the two paths.

Recall that if we parametrize a path C as $(r_1(t),r_2(t),r_3(t))$ with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

$\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt$

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and

$\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}$

The parametrization of the line segment from (2,2,2) to

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)

r'(t) = (-11,4,3)

and

$\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}$

Hence

$\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}$

8 0

2 years ago

Suppose that the function u and w are defined as follows.

mars1129 [50]

That funky circle in the middle is the composition of the function. It asks you to take a function as an input and to yield an output that's another function. It's one of the five function operations, along with adding, subtracting, multiplying, and dividing.

When you compose, you might find the notation w(u(x)) easier to understand. It's saying evaluate u then evaluate w.

For our functions, the compositions are:

u(w(x)) = u(2x²) = -(2x²) - 2 = -2x² - 2

w(u(x)) = w(-x - 2) = 2(-x - 2)² = 2(x² + 4x + 4) = =2x²+ 8x +8

Now we evaluate each composition at 4.

u(w(4)) = -2(4²) - 2 = -2(16) - 2 = -32 -2 = -34

w(u(4)) = -2(2²) +8(2) + 8 = -2(4) + 16 + 8 = -8 + 16 + 8 = 16.

Thus, u(w(4)) = -34 and w(u(4)) = 16.

5 0

2 years ago

<img src="https://tex.z-dn.net/?f=I%20don%27t%20get%20this" id="TexFormula1" title="I don't get this" alt="I don't get this" ali

Sedaia [141]