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2 years ago

What is the average rate of change of y=cos(2x) on the interval 0 pi/2?

Mathematics

2 answers:

Snezhnost [94]2 years ago

7 0

Jesus is the answer to this

icang [17]2 years ago

5 0

Answer:

Average rate of change (A(x)) of y=f(x) over an interval [a, b] is given by:

$A(x) = \frac{f(b)-f(a)}{b-a}$

As per the statement:

Given:

$y=f(x)=\cos (2x)$ and interval $[0, \frac{\pi}{2}]$

At x = 0

$f(0) = \cos (2(0)) = \cos (0) = 1$

At $x = \frac{\pi}{2}$

$f(\frac{\pi}{2}) = \cos (2(\frac{\pi}{2})) = \cos (\pi) =-1$

Substitute the given values in [1] we have;

$A(x) = \frac{f(\frac{\pi}{2})-f(0)}{\frac{\pi}{2}-0}$

⇒ $A(x) = \frac{-1-1}{\frac{\pi}{2}}$

⇒ $A(x) = \frac{-2}{\frac{\pi}{2}}$

⇒ $A(x) = \frac{-4}{\pi}$

Therefore, the average rate of change of y=cos(2x) on the interval [0, pi/2] is, $\frac{-4}{\pi}$

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Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

natima [27]

Answer:

Step-by-step explanation:

The equation of the sphere, centered a the origin is given by $x^2+y^2+z^2 = 64$ . Then, when z=4, we get

$x^2+y^2= 64-16 = 48$ .

This equation corresponds to a circle of radius $4\sqrt[]{3}$ in the x-y plane

c) We will use the previous analysis to define the limits in cartesian and polar coordinates. At first, we now that x varies from $-4\sqrt[]{3}$ up to $4\sqrt[]{3}$ . This is by taking y =0 and seeing the furthest points of x that lay on the circle. Then, we know that y varies from $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , this is again because y must lie in the interior of the circle we found. Finally, we know that z goes from 4 up to the sphere, that is , z goes from 4 up to $\sqrt[]{64-x^2-y^2}$

Then, the triple integral that gives us the volume of D in cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Recall that the cylindrical coordinates are given by $x=r\cos \theta, y = r\sin \theta,z = z$ , where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that $x^2+y^2 = r^2$ . WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ goes from 0 to $2\pi$ . Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

a) For the spherical coordinates, recall that $z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta$ . where $\phi$ is the angle of the vector with the z axis, which varies from 0 up to pi. Note that when z=4, that angle is constant over the boundary of the circle we found previously. On that circle. Let us calculate the angle by taking a point on the circle and using the formula of the angle between two vectors. If z=4 and x=0, then y=4 $\sqrt[]{3}$ if we take the positive square root of 48. So, let us calculate the angle between the vector $a=(0,4\sqrt[]{3},4)$ and the vector b =(0,0,1) which corresponds to the unit vector over the z axis. Let us use the following formula

$\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}$

Therefore, over the circle, $\phi = \frac{\pi}{3}$ . Note that rho varies from the plane z=4, up to the sphere, where rho is 8. Since $z = \rho \cos \phi$ , then over the plane we have that $\rho = \frac{4}{\cos \phi}$ Then, the following is the desired integral

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta$ where the new factor is the jacobian for the spherical coordinates.

d ) Let us use the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$

Note that we can split the integral since the inner part does not depend on theta on any way. If we use the substitution $u = 64-r^2$ then $\frac{-du}{2} = r dr$ , then

$=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}$

3 0

2 years ago

Put the following equation of a line into slope-intercept form, simplifying all

lilavasa [31]

Answer:

y = x -7

Step-by-step explanation:

2x - 2y = 14

The slope intercept form is

y = mx+b where m is the slope and b is the y intercept

Subtract 2x from each side

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

-2y = -2x+14

Divide each side by -2

-2y/-2 = -2x/-2 +14/-2

y = x -7

The slope is 1 and the y intercept is -7

8 0

2 years ago

Read 2 more answers

The table gives estimates of the world population, in millions, from 1750 to 2000. (Round your answers to the nearest million.)

BaLLatris [955]

Answer:

A.) 1508 ; 1870

B.) 2083

C.) 3972

Step-by-step explanation:

General form of an exponential model :

A = A0e^rt

A0 = initial population

A = final population

r = growth rate ; t = time

Using the year 1750 and 1800

Time, t = 1800 - 1750 = 50 years

Initial population = 790

Final population = 980

Let's obtain the growth rate :

980 = 790e^50r

980/790 = e^50r

Take the In of both sides

In(980/790) = 50r

0.2155196 = 50r

r = 0.2155196/50

r = 0.0043103

Using this rate, let predict the population in 1900

t = 1900 - 1750 = 150 years

A = 790e^150*0.0043103

A = 790e^0.6465588

A = 1508.0788 ; 1508 million people

In 1950;

t = 1950 - 1750 = 200

A = 790e^200*0.0043103

A = 790e^0.86206

A = 1870.7467 ; 1870 million people

2.)

Exponential model. For 1800 and 1850

Initial, 1800 = 980

Final, 1850 = 1260

t = 1850 - 1800 = 50

Using the exponential format ; we can obtain the rate :

1260 = 980e^50r

1260/980 = e^50r

Take the In of both sides

In(1260/980) = 50r

0.2513144 = 50r

r = 0.2513144/50

r = 0.0050262

Using the model ; The predicted population in 1950;

In 1950;

t = 1950 - 1800 = 150

A = 980e^150*0.0050262

A = 980e^0.7539432

A = 2082.8571 ; 2083 million people

3.)

1900 1650

1950 2560

t = 1900 - 1950 = 50

Using the exponential format ; we can obtain the rate :

2560 = 1650e^50r

2560/1650 = e^50r

Take the In of both sides

In(2560/1650) = 50r

0.4392319 = 50r

r = 0.4392319/50

r = 0.0087846

Using the model ; The predicted population in 2000;

In 2000;

t = 2000 - 1900 = 100

A = 1650e^100*0.0087846

A = 1650e^0.8784639

A = 3971.8787 ; 3972 million people

3 0

2 years ago

Shania is making a scale diagram of the badminton court at the community center. She uses a scale of 1 centimeter to 0.5 meter t

docker41 [41]

1m = 100cm

1cm → 0.5m

1cm → 0.5 · 100cm = 50cm

The scale: $\dfrac{50}{1}=50$

length = 26.8cm · 50 = 1,340cm = 13.40m

width = 12.2cm · 50 = 610cm = 6.10m

The area of the court:

A = 13.40 · 6.10 = 81.74 m²

8 0

2 years ago

Triangle A has a height of 2.5\text{ cm}2.5 cm2, point, 5, start text, space, c, m, end text and a base of 1.6\text{ cm}1.6 cm1,

konstantin123 [22]

Answer:

Option A

Option D

Option E

Step-by-step explanation:

we know that

If the height and base of triangle B are proportional to the height and base of triangle A

then

Triangle A and Triangle B are similar

Remember that

If two triangles are similar then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

$\frac{h_A}{h_B} =\frac{b_A}{b_B}$

where

h_A and h_B are the height of triangle A and triangle B

b_A and b_B are the base of triangle A and triangle B

In his problem we have

$h_A=2.5\ cm\\b_A=1.6\ cm$

substitute

$\frac{2.5}{h_B} =\frac{1.6}{b_B}$

Rewrite

$\frac{2.5}{1.6} =\frac{h_B}{b_B}$

$\frac{h_B}{b_B}=1.5625$

<u><em>Verify all the options</em></u>

A) we have

$h_B=2.75\ cm\\b_B=1.76\ cm$

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

$\frac{2.75}{1.76}=1.5625$

The ratios are the same

That means that are proportional

therefore

These values could be the height and base of triangle B

B) we have

$h_B=9.25\ cm\\b_B=9.16\ cm$

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

$\frac{9.25}{9.16}=1.0098$

The ratios are not equal

That means that are not proportional

therefore

These values could not be the height and base of triangle B

C) we have

$h_B=3.2\ cm\\b_B=5\ cm$

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

$\frac{3.2}{5}=0.64$

The ratios are not the same

That means that are not proportional

therefore

These values could not be the height and base of triangle B

D) we have

$h_B=1.25\ cm\\b_B=0.8\ cm$

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

$\frac{1.25}{0.8}=1.5625$

The ratios are the same

That means that are proportional

therefore

These values could be the height and base of triangle B

E) we have

$h_B=2\ cm\\b_B=1.28\ cm$

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

$\frac{2}{1.28}=1.5625$

The ratios are the same

That means that are proportional

therefore

These values could be the height and base of triangle B

8 0

2 years ago