Solve the following equation for a. <br> n=b/a+3. <br><br> a=

G(x) = -0.5x^2 + 4x – 2

-BARSIC- [3]

Answer:

Step-by-step explanation:

What is the question?

g(x) = -0.5x² + 4x - 2 is a down-opening parabola. Do you need to know how to put it in vertex form?

vertex (4,6)

focus (4,5.5)

5 0

3 years ago

Name/ Uid:1. In this problem, try to write the equations of the given surface in the specified coordinates.(a) Write an equation

Gemiola [76]

To find:

(a) Equation for the sphere of radius 5 centered at the origin in cylindrical coordinates

(b) Equation for a cylinder of radius 1 centered at the origin and running parallel to the z-axis in spherical coordinates

Solution:

(a) The equation of a sphere with center at (a, b, c) & having a radius 'p' is given in cartesian coordinates as:

$(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=p^{2}$

Here, it is given that the center of the sphere is at origin, i.e., at (0,0,0) & radius of the sphere is 5. That is, here we have,

$a=b=c=0,p=5$

That is, the equation of the sphere in cartesian coordinates is,

$(x-0)^{2}+(y-0)^{2}+(z-0)^{2}=5^{2}$

$\Rightarrow x^{2}+y^{2}+z^{2}=25$

Now, the cylindrical coordinate system is represented by $(r, \theta,z)$

The relation between cartesian and cylindrical coordinates is given by,

$x=rcos\theta,y=rsin\theta,z=z$

$r^{2}=x^{2}+y^{2},tan\theta=\frac{y}{x},z=z$

Thus, the obtained equation of the sphere in cartesian coordinates can be rewritten in cylindrical coordinates as,

$r^{2}+z^{2}=25$

This is the required equation of the given sphere in cylindrical coordinates.

(b) A cylinder is defined by the circle that gives the top and bottom faces or alternatively, the cross section, & it's axis. A cylinder running parallel to the z-axis has an axis that is parallel to the z-axis. The equation of such a cylinder is given by the equation of the circle of cross-section with the assumption that a point in 3 dimension lying on the cylinder has 'x' & 'y' values satisfying the equation of the circle & that 'z' can be any value.

That is, in cartesian coordinates, the equation of a cylinder running parallel to the z-axis having radius 'p' with center at (a, b) is given by,

$(x-a)^{2}+(y-b)^{2}=p^{2}$

Here, it is given that the center is at origin & radius is 1. That is, here, we have, $a=b=0,p=1$ . Then the equation of the cylinder in cartesian coordinates is,

$x^{2}+y^{2}=1$

Now, the spherical coordinate system is represented by $(\rho,\theta,\phi)$

The relation between cartesian and spherical coordinates is given by,

$x=\rho sin\phi cos\theta,y=\rho sin\phi sin\theta, z= \rho cos\phi$

Thus, the equation of the cylinder can be rewritten in spherical coordinates as,

$(\rho sin\phi cos\theta)^{2}+(\rho sin\phi sin\theta)^{2}=1$

$\Rightarrow \rho^{2} sin^{2}\phi cos^{2}\theta+\rho^{2} sin^{2}\phi sin^{2}\theta=1$

$\Rightarrow \rho^{2} sin^{2}\phi (cos^{2}\theta+sin^{2}\theta)=1$

$\Rightarrow \rho^{2} sin^{2}\phi=1$ (As $sin^{2}\theta+cos^{2}\theta=1$ )

Note that $\rho$ represents the distance of a point from the origin, which is always positive. $\phi$ represents the angle made by the line segment joining the point with z-axis. The range of $\phi$ is given as $0\leq \phi\leq \pi$ . We know that in this range the sine function is positive. Thus, we can say that $sin\phi$ is always positive.

Thus, we can square root both sides and only consider the positive root as,

$\Rightarrow \rho sin\phi=1$

This is the required equation of the cylinder in spherical coordinates.

Final answer:

(a) The equation of the given sphere in cylindrical coordinates is $r^{2}+z^{2}=25$

(b) The equation of the given cylinder in spherical coordinates is $\rho sin\phi=1$

7 0

3 years ago

The differential equation y′′=0 has one of the following two parameter families as its general solution: yyyy=C1ex+C2e−x=C1cos(x

svet-max [94.6K]

Answer:

y_c = 2 + 10*x

Step-by-step explanation:

Given:

y'' = 0

Find:

- The solution to ODE such that y(0) = 2, y'(0) = 10

Solution:

- Assuming a solution y = Ce^(mt)

So, y' = C*me^(mt)

y'' = C*m^2e^(mt)

- Back substitute into given ODE, we get:

y'' = C*m^2e^(mt) = 0

e^(mt) can not be equal to zero

- Hence, m^2 = 0

m = 0 , 0 - (repeated roots)

- The complimentary function for repeated roots is:

y_c = (C1 + C2*x)*e^(m*t)

y_c = C1 + C2*x

- Evaluate @ y(0) = 2

2 = C1 + C2*0

C1 = 2

-Evaluate @ y'(0) = 10

y'(t) = C2 = 10

Hence, y_c = 2 + 10*x

5 0

3 years ago

Y = (x²+3)(2x-6)<br> What is the derivative?

ahrayia [7]

Answer:

y = (x²+3)(2x-6)

dy/dx=d(x²+3)(2x-6)/dt=(x²+3)d(2x-6)/dt+(2x-6)d(x²+3)/dt= (x²+3)(2)+(2x-6)(2x)=2x²+6+4x²-12x=6x²-12x+6=6(x²-2x+1)

4 0

3 years ago

If f(x)= 1/x+2 and a>3, which of the following could be f(a)

Xelga [282]

I think it’s D not sure

5 0

3 years ago

Solve the following equation for a. n=b/a+3. a=