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BabaBlast [244]
2 years ago
15

Rewrite the parametric equations in Cartesian form: x(t)=4+t, y(t)=2sqrt(t). y=?

Mathematics
1 answer:
Dmitriy789 [7]2 years ago
6 0

Answer:

Both of these curves are equal to each other.

Step-by-step explanation:

We have

x(t) = 4 + t

and

y(t) = 2 \sqrt{t}

Solve for t in the x function

x = 4 + t

Subsitue this in for t.

y = 2 \sqrt{x - 4}

We get

a square root function shifted to right 4 units and stretched vertically by a factor of 2.

We can also prove this by graphing.

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Find the average value of f over region
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The area of D is given by:

\int\limits \int\limits {1} \, dA = \int\limits_0^7 \int\limits_0^{x^2} {1} \, dydx  \\  \\ = \int\limits^7_0 {x^2} \, dx =\left. \frac{x^3}{3} \right|_0^7= \frac{343}{3}

The average value of f over D is given by:

\frac{1}{ \frac{343}{3} }  \int\limits^7_0  \int\limits^{x^2}_0 {4x\sin(y)} \, dydx  = -\frac{3}{343}  \int\limits^7_0 {4x\cos(x^2)} \, dx  \\  \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49
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3 years ago
-2x + 3x - 5 = 6 - 21<br><br> FIND x
trasher [3.6K]

Before we do any solving, we need to simplify both sides.

On the left, we can simplify -2x + 3x - 5 to x - 5.

On the right, we can simplify 6 - 21 to -16.

So we have x - 5 = -16.

Solving from here, we add 5 to both sides to get x = -11.

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3 years ago
Which image is a reflection of the orange letter K in Quadrant I?
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Answer: Quadrant 4

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
Use the surface integral in​ Stokes' Theorem to calculate the circulation of the field Bold Upper F equals x squared Bold i plus
Alinara [238K]

Answer:

The circulation of the field f(x) over curve C is Zero

Step-by-step explanation:

The function f(x)=(x^{2},4x,z^{2}) and curve C is ellipse of equation

16x^{2} + 4y^{2} = 3

Theory: Stokes Theorem is given by:

I= \int \int\limits {{Curl f\cdot \hat{N }} \, dx

Where, Curl f(x) = \left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right]

Also, f(x) = (F1,F2,F3)

\hat{N} = grad(g(x))

Using Stokes Theorem,

Surface is given by g(x) = 16x^{2} + 4y^{2} - 3

Therefore, tex]\hat{N} = grad(g(x))[/tex]

\hat{N} = grad(16x^{2} + 4y^{2} - 3)

\hat{N} = (32x,8y,0)

Now,  f(x)=(x^{2},4x,z^{2})

Curl f(x) = \left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right]

Curl f(x) = \left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\x^{2}&4x&z^{2}\end{array}\right]

Curl f(x) = (0,0,4)

Putting all values in Stokes Theorem,

I= \int \int\limits {Curl f\cdot \hat{N} } \, dx

I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx

I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx

I=0

Thus, The circulation of the field f(x) over curve C is Zero

3 0
3 years ago
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