How do you solve the following Calculus questions? Please help :)?

Question

Hi, I am having trouble with a few questions from my Calculus packet...any replies will be appreciated.

#7 C

#8 A, B, C (I just don't really understand what the standard forms are)

#11 A

#16 B

#31 C

#36 A & D

http://www.math.unb.ca/ready/paper.pdf

Thank you mucho!

Answer 1

7.

A = 2pi*r^2 + 2pi*rh

0 = 2pi*r^2 + 2pi*rh-A

now use the quadratic formula, and solve for r=[-b+sqrt(b^2-4ac)]/2a (the only case where r can be positive)

8.  The standard forms are those used for graphing conic sections.  I'm not trying to discourage you, but that is actually a huge big deal.  If  you don't know those (or are completely lost) you need to re-take precalc or at least study analytical geometry pretty hard.  I'll solve the first one for you.

A.

y=x^2 + 4x + 3

y-3 = x^2 + 4x

now take 1/2 the coefficient of the x-term, then square that result, and add it to both sides:  (4/2)^2 = 4

y-3+4 = x^2 + 4x + 4

by doing this, you forced the right hand side to be a perfect square

(y+1) = (x+2)^2

this is the graph of a parabola, which opens up, with a vertex of (-2,-1)

the vertex is always the opposite sign of the numbers in parentheses, once you get the polynomial in standard form

11. 3(sinx)^2 = (cosx)^2

(sinx/cosx)^2 = 3

sinx/cosx = sqrt(3)

tan x = sqrt 3

x = 60 degrees

16.  12x^3 + 8x^2 - x - 1 = 0

by the rational root theorem, all roots must be of the form of +/-(p/q), where p is a factor of the constant term (the -1) and q is a factor of the coefficient of the highest order variable (the 12).

So all possible roots are +/- of each of these:

1/1, 1/2, 1/3, 1/4, 1/6, 1/12

using synthetic division (look it up if you need to), x=1/2 is a root, so then (x-1/2) is a factor:

(x-1/2)(12x^2+2x-2)

the quadratic formula gives us the factorization of the rest:

(x-1/2)(x+1/3)(x+1/2)    (foil it out to check the work)

31 C.

x=sin t

y=cos t

solve for t:  t = sin-1 (x)  [this is "sine inverse of x"]

now substitute in for y:

y = cos (sin-1(x))

36. A.  Just let y = -x

cos(x+y) = cosx * cosy - sinx * siny

cos(x-x) = cosx * cos(-x) - sinx * sin(-x)

(since cos(-x) = cos(x) and sin(-x) = -sin(x)

cos (0) = cosx * cosx + sinx * sin x

1 = (cosx)^2 + (sinx)^2

D.  Let y = x:

cos(x+y) = cosx * cosy - sinx * siny

cos(x+x) = cosx * cosx - sinx * sinx

cos(2x) = (cosx)^2 - (sinx)^2

but from 36A, 1 = (cosx)^2 + (sinx)^2

or -(sinx)^2 = (cosx)^2 - 1, now substitute this in for - (sinx)^2

cos(2x) = (cosx)^2 + (cosx)^2 - 1

cos(2x) = 2(cosx)^2 - 1

Good luck!

Answer 2

WTF!

lol

I hope this reply would be appreciated..

Your welcome

Answer 3

I won't give you the answers to your homework, but I'll see what I can do to help you figure out the ones I know.

For #11A, you need to remember your trig identities, particularly that sin²x+cos²x=1. So convert the sin²x into 1-cos²x to get:

3(1-cos²x)=cos²x

Then solve, remembering your inverse trig functions.

For #31C, it's just a unit circle. Don't let the swapped sin/cos fool you; it' still a circle, it just "begins" at the top instead of the right. But it's a closed, sideless, figure with no endpoints, so it doesn't matter.

For #36D, use identity (C) and identity (a). If in (C) x=y, then you now have:

cos(2x)=cos²x-sin²x

And according to identity (a):

sin²x=1-cos²x

So:

cos(2x)=cos²x-(1-cos²x)

cos(2x)=2cos²x - 1

Sorry I couldn't figure out the rest.

-IMP ;) :)  

Answer 4

Those are not calculus questions. Some of them are even earlier than pre-calc. The packet is labelled "Are you ready for Calculus?" for a reason. If you are having trouble with that many questions, maybe you should retake pre-calculus first.