1. If the 20 inch (radius) blades of a ceiling fan are rotating at the rate of 4 revolutions per second, what is the linear speed (in ft/min) if the outer tip of the blades??

2. Right triangle. opp side=7, adj=10, hyp=x. Theta is between 7 and x, beta is between x and 10.

Find: theta, beta, x, sin theta, cos beta, tan theta, sec theta, csc beta, cot theta.

and using exact values: show that cos@^2 + sin@^2=1

3. 27 inch (diameter) wheels on a bike are rotating 5pi radians per second.
-what is the equivlaent rate in revolutions per minute??

-linear speed of the bike in mph?

-if the bike is pulling a little kid in a cart with 5 inch radius wheels, at how many revs per minute at the 5 inch wheels turning?

4. Show a detailed graph of y=3sin(1.5x-(pi/2))+2 for 0<x<2pi
- determine first two x intercepts
-determine first relative max pt.
-determine rel min pt after x=3
-what is the amplitude?
show manual steps to determine period, and what is the phase shift

5. theta=arctan(x/9)=tan^-1(x/9) complete the labels of the 3 sides of the right traingle and determine sin theta and cos theta in terms of x. (x is given opposite lower corner of right angle)

6. what angle (degrees and radians) is measured by a 2 inch arc ona circle with radius of 1/2 inch.

sec25degrees=csc(___) ?= 1/cos(___) ? = ?

8. assume the moon orbits the earth in a circular way every 27 days witht he earth being the center of the circular orbit. the moon is aprx 240,000 miles from the center of the earth. at how many mph is the moon moving???

I would very much appriciate if someone tackled some of these problems. I have done some of them but would like to see other peoples answers. THANKS

2. Originally Posted by Stuck686
1. If the 20 inch (radius) blades of a ceiling fan are rotating at the rate of 4 revolutions per second, what is the linear speed (in ft/min) if the outer tip of the blades??
v = r*(omega)
where v is the linear speed, r is the radius, and (omega) is the angular speed.

We know that
(omega) = 4 rev/s = 4 * 2(pi) rad/s = 8(pi) rad/s

So
v = (20 ft)*( 8(pi) rad/s ) = 160(pi) ft/s = 502.655 ft/s.

-Dan

3. Originally Posted by Stuck686
2. Right triangle. opp side=7, adj=10, hyp=x. Theta is between 7 and x, beta is between x and 10.

Find: theta, beta, x, sin theta, cos beta, tan theta, sec theta, csc beta, cot theta.

and using exact values: show that cos@^2 + sin@^2=1
If (theta) is between the sides 7 and x and 7 is the "opposite" side then (theta) is the angle across from the side 10. So:
tan(theta) = 10/7 ==> (theta) = atn(10/7) = 0.96007 rad.

Similarly
tan(beta) = 7/10 ==> (beta) = atn(7/10) = 0.610726 rad

As a quick check, these angles should add up to be a right angle, or (pi)/2 rad = 1.5708 rad. You can check to see that they do.

Now, by the Pythagorean Theorem
7^2 + 10^2 = x^2

x = sqrt{149} = 12.2066

sin(theta) = 10/x = 10/sqrt{149} = 0.819232

cos(beta) = 10/x = 10/sqrt{149} = 0.819232 <-- Yes, sin(theta) = cos(beta)

tan(theta) = 10/7 = 0.142857 <-- as before

sec(theta) = 1/cos(theta) = 1/(7/x) = x/7 = sqrt{149}/7

csc(beta) = 1/sin(beta) = 1/(7/x) = x/7 = sqrt{149}/7 (Yes, sec(theta) = csc(beta).)

cot(theta) = 1/tan(theta) = 1/(10/7) = 7/10

Now we wish to show that sin^2(@) + cos^2(@) = 1 by example. We have two angles to work with here:
sin^2(theta) + cos^2(theta) = [10/sqrt{149}]^2 + [7/sqrt{149}]^2 = 100/149 + 49/149 = 1 (Check!)

sin^2(beta) + cos^2(beta) = [7/sqrt{149}]^2 + [10/sqrt{149}]^2 = 49/149 + 100/149 = 1 (Check!)

-Dan

4. Originally Posted by Stuck686
3. 27 inch (diameter) wheels on a bike are rotating 5pi radians per second.
-what is the equivlaent rate in revolutions per minute??

-linear speed of the bike in mph?

-if the bike is pulling a little kid in a cart with 5 inch radius wheels, at how many revs per minute at the 5 inch wheels turning?
The diameter is 27 inches, so the radius is r = 13.5 inches.

(omega) = 5(pi) rad/s = 5(pi)/[2(pi)] rev/s = 2.5 rev/s

The linear speed of the bike may be found by:
v = r*(omega) = (13.5 inches)*( 5(pi) rad/s ) = 212.058 in/s.

We want this in mph. There are 17.6 in/s in 1 mi/h, so the speed is:
(212.058/17.6) mi/h = 12.0487 mph

For the cart, r = 5 in and again
v = r*(omega)

(omega) = v/r = (212.058 in/s)/(5 in) = 42.4115 rad/s

-Dan

5. Originally Posted by Stuck686
4. Show a detailed graph of y=3sin(1.5x-(pi/2))+2 for 0<x<2pi
- determine first two x intercepts
-determine first relative max pt.
-determine rel min pt after x=3
-what is the amplitude?
show manual steps to determine period, and what is the phase shift
I have attached a graph at the bottom of the post.

A zero of the function can be found at:
3sin(1.5x-(pi)/2)+2 = 0

3sin(1.5x-(pi)/2) = -2

sin(1.5x - (pi)/2) = -2/3

1.5x - (pi)/2 = asn(-2/3) = -0.729728 (rad)

1.5x = (pi)/2 - 0.729728 = 0.841069 (rad)

so we can see this is the first solution. How do we get the second?

Well, this sine function is symmetric about the line x = 2(pi)/3. So the distance between the first x solution and the line of symmetry is 2(pi)/3 - 0.17848(pi) = 0.488186(pi). Thus the next zero is this much past the line x = 2(pi)/3. Or the second zero is at 2(pi)/3 + 0.488186(pi) = 1.15485(pi).

So the first two zeros are at
x = 1.15485(pi)

(How did I get the line x = 2(pi)/3 as the line of symmetry? Well, sin(x) is symmetric about x = (pi)/2. Now we need to apply the transformation: (x + (pi)/2)/1.5, which is the reverse of the transformation 15.x - (pi)/2. This gives ( (pi)/2 + (pi)/2 )/1.5 = (pi)/1.5 = 2(pi)/3 as the line of symmetry.)

The first relative max point can be found in the same manner as I found the line of symmetry. The function sin(x) has its first relative max at x = (pi)/2, which is our line of symmetry. Thus we know that the first relative max will appear at x = 2(pi)/3, which is the line of symmetry for our sine function.

Similarly the first relative minimum point after x = 3 can be found by considering the sin(x) function: The minima of the sin(x) function appears at x = 3(pi)/2. Applying the transformation:
x = ( 3(pi)/2 + (pi)/2 )/1.5 = ( 2(pi) )/1.5 = 4(pi)/3 > 3

For the last two parts consider the equations:
y=3sin(1.5x-(pi/2))+2
and
y = Asin(kx + (phi) ) + B

A is the amplitude, k is the wavenumber, (phi) is the phase, and B is the vertical displacement.

So the amplitude is 3 and the phase shift is -(pi)/2.

To get the period (which is the same as the wavelength in this case) consider that the wavenumber k = 1.5. Thus
k = 2(pi)/(lambda) <-- (lambda) is the wavelength, ie. period

(lambda) = 2(pi)/k = 2(pi)/1.5 = 4(pi)/3

-Dan

6. aweomse thanks, could someone also do 8 i dont think i get that

7. Originally Posted by Stuck686
8. assume the moon orbits the earth in a circular way every 27 days witht he earth being the center of the circular orbit. the moon is aprx 240,000 miles from the center of the earth. at how many mph is the moon moving???
The moon orbits in a circle with radius 240 000 miles. it makes one complete revolution in 27 days.

now speed = distance/time

what is the distance? the circumference of the circle
what is the time? 27 days = 648 hours

distance = 2pi*r = 2pi*(240 000) = 480 000*pi miles

=> speed = (480 000 pi)/648 = 20000*pi/27 mph ~= 2327.11 mph

8. Originally Posted by Stuck686
5. theta=arctan(x/9)=tan^-1(x/9) complete the labels of the 3 sides of the right traingle and determine sin theta and cos theta in terms of x. (x is given opposite lower corner of right angle)
theta = atn(x/9)
So imagine a right triangle with theta in the bottom left corner. We know that
tan(theta) = x/9
so we know that x is the side opposite to theta and 9 is adjacent. Thus we know the hypotenuse, h, is:
h = sqrt{x^2 + 9}

So
sin(theta) = x/sqrt{x^2 + 9}
cos(theta) = 9/sqrt{x^2 + 9}

-Dan

9. Originally Posted by Stuck686
6. what angle (degrees and radians) is measured by a 2 inch arc ona circle with radius of 1/2 inch.
An angle measured by the arc of a circle of radius r is given (in radians) as:
s = r*(theta)
where s is the arc length.

So.
s = 2 in and r = 1/2 in.
Thus
(theta) = s/r = (2 in)/(1/2 in) = 4 rad.

What is this in degrees? Well (pi) rad = 180 degrees, so:

-Dan

10. Originally Posted by Stuck686

sec25degrees=csc(___) ?= 1/cos(___) ? = ?
tan(-4(pi)/3)

Let's hold off on that for a moment.
The angle 4(pi)/3 rad is greater than (pi) rad and less than 3(pi)/2 rad. So it is in Quadrant III. This means the reference angle is 4(pi)/3 rad - (pi) rad = (pi)/3 rad and is in QIII. Be careful, though. The original angle is negative, so the actual angle is a reference angle of (pi)/3 rad in QII. (Sketch a picture if you find this confusing.)

So the reference angle is (pi)/3 rad and in QII.

Now, what is tan(-4(pi)/3)?

tan(-4(pi)/3) = -tan((pi)/3) <-- Negative because the angle is in QII.

= -sqrt{3}

As to cos(500) we do that the same way:
cos(500) = cos(500 - 360) <-- We can always subtract 360 degrees

= cos(140)
140 degrees is in QII with a reference angle of 180 degrees - 140 degrees = 40 degrees. As cosine is negative in QII we get:
cos(500) = cos(140) = -cos(40)

As to sec(25), note that sec(25) = 1/cos(25). We wish to find the cosecant of an angle (theta) such that csc(theta) = sec(25). Well, csc(theta) = 1/sin(theta). Can we find a (theta) such that sin(theta) = cos(25)?

Yup! Take a look at the right triangle where one of the angles is 25 degrees. The other angle will be 65 degrees. Note that sin(65) = cos(25). (Construct the triangle and apply the definitions.) Thus:
sec(15) = 1/cos(25) = 1/sin(65) = csc(65)

-Dan

11. Originally Posted by topsquark

For the cart, r = 5 in and again
v = r*(omega)

(omega) = v/r = (212.058 in/s)/(5 in) = 42.4115 rad/s

-Dan

can you check something for me? if you solve the rad/s for revolutions per minute, do you get 3997.189rev/m ?

12. Originally Posted by Stuck686
can you check something for me? if you solve the rad/s for revolutions per minute, do you get 3997.189rev/m ?
(42.4115 rad/s) x (1 rev/ 2(pi) rad) x (60 s/1 min) = 405 rpm

-Dan