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I need help in Calc

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JPRaup

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Im taking AP Calc next year as a senior, we need to do a summer project thing for it, basically like a bunch of worksheets. So today is the last day of summer and Im only about 1/4 done. I threw away all my precalc notes and need help :(

ok here is something I forget

if

f(x) = {(3,5), (2,4), (1,7)} and h(x) = {(3,2), (4,3), (1,6)}

what is

(f + h)(1) =
(f 0 h)(3) =
f^-1 (x) =
1/f(x) =

I forget how to do all this stuff....

thanks in advance
 
another question

how do you simplify


(x^(3/2)) * (x + x^(5/2) - x^2)

and how do you eliminate the parameter and write a rectangular equation for:

x=t^2 +3
y=2t
 
JPRaup said:
another question

how do you simplify


(x^(3/2)) * (x + x^(5/2) - x^2)

and how do you eliminate the parameter and write a rectangular equation for:

x=t^2 +3
y=2t

plenary-shocked.jpg
 
You're taking AP calc next year and they've got you working with parametric equations?

That's not on the AP test, AFAIK. I'm just worknig with these now and I'm in calc 3.

And the answer to the one about the parametric equation (as best as I guess you mean) is:
x^(5/2)+x^(4)-x^(7/2)
 
Ecrofirt said:
You're taking AP calc next year and they've got you working with parametric equations?

That's not on the AP test, AFAIK. I'm just worknig with these now and I'm in calc 3.

i guess so, ahh, im fucked, i have 2.5 hours to do the rest of this before i go to work, then school is tommorow :(

can u help me with my first post

and what is ln 1/2
 
JPRaup said:
another question

how do you simplify


(x^(3/2)) * (x + x^(5/2) - x^2)

and how do you eliminate the parameter and write a rectangular equation for:

x=t^2 +3
y=2t
For the first part, just distribute the x^(3/2), and when you multiply exponents like this you add them together. So, to add 3/2 to the first one, you'll have to have a common denominator. The x is really x^1, so to get a common denominator it would be x^(2/2). The second one already has a common denominator, and the third one would become x^(4/2). One they're all added together, you get: x^(5/2) + x^4 - x^(7/2). Not sure what to do from there, I can't think of a way to factor it.

For the second part, in the x=t^2 + 3, you get t by itself. So that becomes (+/-)square root of (x - 3) = t. Then you substitute what t is equal to into the second equation. This becomes y= 2(plus and minus square root of x-3).
 
Sorry man, that stuff from the first post is slippnig my mind.

Do you have a calculator? ln(1/2) is -ln(2) or -.693147
 
Ecrofirt said:
Sorry man, that stuff from the first post is slippnig my mind.

Do you have a calculator? ln(1/2) is -ln(2) or -.693147

yep got a ti-83 but i got that one, and thank u soo much for explaining the other one to me, ok now more calc
 
I'm not 100% sure, but I think the multiplication takes precedence over the factorial, so...

if n = 3

3(n+1)! / 5n!

is equivalent to

3(3+1)! / 5(3)!

is equivalent to

12! / 15!

is equivalent to

(12x11x10x9x8x7x6x5x4x3x2x1) / (15x14x13x12x11x10x9x8x7x6x5x4x3x2x1)

is equivalent to

1 / (15x14x13)
 
Oh man, if this is calculus then I'm screwed, I forgot how to do all this shit, but I'm expecting grade 12 calculus to be easier, hopefully.
 
thanks for the help
now i need to determine the exact value of thes

1. sin 0
2. sin pi/2
3. sin 3pi/4
4. cos pi
5.cos 7pi/6
6. cos pi/3
7. tan 7pi/4
8. tan pi/6
9. tan 2pi/3
10. tan pi/2
11. cos((sin ^-1) 1/2)
12. sin ^-1 (sin 7pi/6)


ill try to edit the thread as a figure them out, but any help is appreciated
 
Meatpuppet said:
I'm not 100% sure, but I think the multiplication takes precedence over the factorial, so...

if n = 3

3(n+1)! / 5n!

is equivalent to

3(3+1)! / 5(3)!

is equivalent to

12! / 15!

is equivalent to

(12x11x10x9x8x7x6x5x4x3x2x1) / (15x14x13x12x11x10x9x8x7x6x5x4x3x2x1)

is equivalent to

1 / (15x14x13)

I WAS WRONG!

Sorry, factorial takes precedence over multiplication, should be:

3(n+1)! / 5n!

is equivalent to

3(n+1)n! / 5n!

is equivalent to

3(n + 1) / 5

is equivalent to

.6n + .6

Sorry, I was a bit hazy on the order of precdence outside of the bog standard operators. Mea culpa
 
Ecrofirt said:
You're taking AP calc next year and they've got you working with parametric equations?

That's not on the AP test, AFAIK. I'm just worknig with these now and I'm in calc 3.

And the answer to the one about the parametric equation (as best as I guess you mean) is:
x^(5/2)+x^(4)-x^(7/2)

Yes it is. I took the AP Calc BC test last year, and parametric equations were on it.

Unless you're talking about Calc AB.
 
1. sin 0 :0
2. sin pi/2 :1
3. sin 3pi/4 :1/square root of 2
4. cos pi : -1
5.cos 7pi/6 : negative square root of 3/2
6. cos pi/3 : 1/3
7. tan 7pi/4 : 1
8. tan pi/6 : 1/square root of 3
9. tan 2pi/3 : negative square root of 3
10. tan pi/2 : you have to divide by zero, no answer
11. cos((sin ^-1) 1/2) : dunno
12. sin ^-1 (sin 7pi/6) : dunno
 
Litigation Manuel said:
1. sin 0 :0
2. sin pi/2 :1
3. sin 3pi/4 :1/square root of 2
4. cos pi : -1
5.cos 7pi/6 : negative square root of 3/2
6. cos pi/3 : 1/3
7. tan 7pi/4 : 1
8. tan pi/6 : 1/square root of 3
9. tan 2pi/3 : negative square root of 3
10. tan pi/2 : you have to divide by zero, no answer
11. cos((sin ^-1) 1/2) : dunno
12. sin ^-1 (sin 7pi/6) : dunno


for #11:

arcsin(1/2) = sin^-1 (1/2)

cos(arcsin(1/2)) = cos (pi/6) = (square root of 3) / 2




For #12: sin 7pi/6 is -1/2

arcsin(-1/2) = 11pi/6. (although I think that the correct answer, technically, is -pi/6)
 
GaimeGuy said:
Yes it is. I took the AP Calc BC test last year, and parametric equations were on it.

Unless you're talking about Calc AB.


nope taking AB

thanks for the help guys

next part

doman and range for the following


y = square root of (x-4)
y = square root of ((x^2)-4)
y = square root of (4 -(x^2))
y = square root of ((x^2) + 4)
 
Dude. do your own work, man. It's fine asking help for concepts, but you're asking us to do your work for you.

Here's a hint: What can't you perform a square root operation on? What numbers have no real roots?
 
GaimeGuy said:
Dude. do your own work, man. It's fine asking help for concepts, but you're asking us to do your work for you.

Here's a hint: What can't you perform a square root operation on? What numbers have no real roots?


ha, im not even asking for like 1/5 of the answers, there are soo many problems. I know most of them, but Im asking for help on the ones where i dont even know where to start
 
Ok. I'll tell you what to do to find the domains. You can't perform a square root on a negative. let the stuff that you're rooting be greater than or equal to zero, and solve for x.

ex: domain of y=square root of (x-4)

x-4 must be greater than or equal to 0

x must be greater than or equal to 4.

x = [4,∞)
There's your domain.
 
GaimeGuy said:
Ok. I'll tell you what to do to find the domains. You can't perform a square root on a negative. let the stuff that you're rooting be greater than or equal to zero, and solve for x.

ex: domain of y=square root of (x-4)

x-4 must be greater than or equal to 0

x must be greater than or equal to 4.

x = [4,∞)
There's your domain.

thank you i got all those, ill let you know if I need help with any more, Im almost finished, i really appreciate all the help
 
JPRaup said:
thanks for the help
now i need to determine the exact value of thes

1. sin 0
2. sin pi/2
3. sin 3pi/4
4. cos pi
5.cos 7pi/6
6. cos pi/3
7. tan 7pi/4
8. tan pi/6
9. tan 2pi/3
10. tan pi/2
11. cos((sin ^-1) 1/2)
12. sin ^-1 (sin 7pi/6)


ill try to edit the thread as a figure them out, but any help is appreciated

Do you need to know how to evaluate their specific values, or just how to come across them?

Basically, these are the simplest trig identities. Multiples of 90 degrees will have values of either 0 or +1/-1

Sin 0 = 0
Sin 90 = 1
Sin 180 = 0
Sin 270 = -1

Cos 0 = 1
Cos 90 = 0
Cos 180 = -1
Cos 270 = 0

Sin 30 = .5
Sin 60 = .5sqrt(3)
Sin 45 = sqrt(.5)

Cos 30 = sqrt(.5)
Cos 60 = .5
Cos 45 = sqrt(.5)

tan 0 = 0
tan 30 = 1/sqrt(3)
tan 45 = 1
tan 60 = sqrt(3)
tan 90 = undefined

and so on, bearing in mind the quadrant rule:

In the first quadrant of a circle (0.0 -90 degrees, top right quarter), sin cos and tan are all positive.
In the second quadrant of a circle (90.0 -180 degrees, top left quarter), sin alone is positive.
In the third quadrant of a circle (180.0 - 270 degrees, bottom left quarter), tan alone is positive.
In the fourth quadrant of a circle (270.0 - 0 degrees, bottom right quarter), cos alone is positive.

360 degrees is 2pi radians, making 90 degrees pi/2, 60 degrees pi/3 etc etc

sin^-1 works in the opposite sense, in the way that log works in the opposite sense of exponential:

sin pi/2 = sin 90 = 1

therefor

sin^-1 = pi/2 = 90

therefor

1. sin 0 = 0
2. sin pi/2 = sin 90 = 1
3. sin 3pi/4 = sin 135 = sqrt(.5) (sin 45 in the second quadrant)
4. cos pi = cos 180 = -1
5.cos 7pi/6 = cos 210 = -.5sqrt(3) (cos 20 in the third quadrant)
6. cos pi/3 = cos 60 = .5
7. tan 7pi/4 = tan 315 = -1 (tan 45 in the fourth quadrant)
8. tan pi/6 = tan 30 = 1/sqrt(3)
9. tan 2pi/3 = tan 120 = - 1/sqrt(3) (tan 30 in the second quadrant)
10. tan pi/2 = tan 90 = undefined (infinity)
11. cos((sin ^-1) 1/2) = cos pi/6 = cos 30 = .5sqrt(3)
12. sin ^-1 (sin 7pi/6) = 7piu/6 = 210 degrees
 
GaimeGuy said:
Ok. I'll tell you what to do to find the domains. You can't perform a square root on a negative. let the stuff that you're rooting be greater than or equal to zero, and solve for x.

ex: domain of y=square root of (x-4)

x-4 must be greater than or equal to 0

x must be greater than or equal to 4.

x = [4,∞)
There's your domain.

Imaginary numbers to the rescue!

i (or j if you're one of those electrical engineering tools) for the win!

Also, no offense man, but if you're having this much trouble with these questions, I fear for you once term starts.
 
Meatpuppet said:
Imaginary numbers to the rescue!

i (or j if you're one of those electrical engineering tools) for the win!

Also, no offense man, but if you're having this much trouble with these questions, I fear for you once term starts.

no, its all slowly coming back, ive done nothing this whole summer. I understand it now.
 
Meatpuppet said:
Imaginary numbers to the rescue!

i (or j if you're one of those electrical engineering tools) for the win!

Also, no offense man, but if you're having this much trouble with these questions, I fear for you once term starts.

of course imaginary numbers could work. But imaginary numibers aren't used in Calc AB or BC. :)
 
Scary thread! Haha, math stuff makes me feel like the biggest dumbass in the world.

GL with your class, that's all I can assist you with..
 
JPRaup said:
ok im almost done, i didnt get this one though

solve for y

ln y = 2t -3

I think you've got to raise e to both sides, like so:
e^lny=e^(2t-3)
that gives you y=e^(2t-3)

Don't take my word for it though, as I'm getting an implicit differentation vibe from this, or something like that.
 
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