Points of Inflexion and Trigonometric Curves (more maths help...)

Flay

I've got the basics of 'Intergration Using Substitution' down, but I'm having some trouble with the more complex questions. I was hoping someone might be able to go through one of the questions step-by-step so that I can understand it better.

So here's the question;

Use the substitution u = 5 - x to find ∫x√(5 - x) dx.

Can anyone help?

TechBoy

With u-substitution your goal is to take a complicated function and make it simpler by substituting in the variable U where you previously had a trickier equation with X.

In this problem your U is given to you, so that's handy. U = 5 - x.

We substitute that in and the equation becomes: ∫x√u dx. Before we can integrate with respect to U, we need to convert all X's to U's, so we need to take care of dx as well as that lone X.

dx is easy enough, if we take the derivative of the equation U = 5- x, we find that du = -dx. So that's taken care of, dx = -du.

But what about that lone X? Well, we do have the equation U = 5-x, maybe we could solve for X in terms of U...?

TheGreat2nd

Solution inside

u = 5 - x
du = -dx
dx = -du

x = 5 - u

∫x√(5 - x) dx
∫(5-u)(u^1/2) du
∫5u^1/2 - u^3/2 du

EDIT: Aka, what TechBoy said.
But, remember that you can also solve for X in terms of U.

corcorigan

Substitution is awesome. Turn something horrible into something comically easy, and wow people.

Flay

Thanks guys. It seems my working was fine up until about halfway through the question, but I was buggering up the substitution. :P

KingAgamemnon

A good rule of thumb with U substitution.

Make U equal to what is under a radical or in a denominator or in the argument of a trig. function.

85% of first semester integrals can be solved with that method.

Flay

Looks like I'm stuck again... Here's the problem;

Find the volume of the solid of revolution formed when the curve y = x(x^3 - 3)^2 is rotated about the x-axis from x = 0 to x = 1.

Big Dookie

Alright, it's been a while since I've done this, so bear with me. I'll try to explain it the best I can remember it. If I was on a computer with MATLAB I could explain this much better with some plots, but I'll try to just explain it in a way you can visualize.

When you have a function that is rotated about the X-Axis, you basically have a series of concentric disks that are stacked sideways on top of one another. It's like those 3D puzzles, where the pieces are flat and seemingly random shaped, but when you stack them in the correct order, they create some kind of 3 Dimensional figure like a statue or a building. Now, visualize each of these pieces as being perfectly round (since you're simply rotating in a circle around the x-axis), and infinitely thin. If you could "add up" all of these disks, you end up with the volume of the shape. That's basically the conceptual basis of the problem.

If you want to know the exact solution, click below:

Now, since each of the disks is a perfect circle, the area of each disk is simply A = pi(r)^2, right? Easy enough. So now we just need to know the radius of all the disks. Thankfully, this is given to us in the problem - it's simply the function itself. The radius of any given disk is y, which in this case is equal to x(x^3 - 3)^2. Therefore, the area of any given disk becomes:

A = pi[x(x^3 - 3)^2]^2

Which turns out to be a heck of a polynomial. This probably the most difficult part of the problem, honestly. If I expanded it correctly, it becomes:

A = pi(x^14 - 12(x^11) - 54(x^5) + 81(x^2))

Check me on that, because if I messed up, it was probably there.

So now you have the Area of each individual disk, and so the final step is add up all the disks together. This is basically an integration of the Areas above from your limits of integration, which are x=0 to x=1.

V = ∫(A)dx integrated from 0 to 1. Pull the pi constant out in front, and the integral is:

V = pi ∫ (x^14 - 12(x^11) - 54(x^5) + 81(x^2)) dx | 0 to 1

V = pi (1/15(x^15) - x^12 - 9(x^6) + 27(x^3)) | 0 to 1

Which becomes:

V = (256*pi) / 15

I hope I did that right.

KingAgamemnon

Big Dookie is spot on. I was worried it may have been a doughbut shape, but the graph comes back looking pretty solid (almost like a tan graph, around the origin), plus it was around the x axis.

Anyway, you've kept me from digging up my calc. book, gg BD.

Big Dookie

Hah, yes, the ring-shaped figures are a bit more difficult. Not too terrible, but certainly worse than the one posted. Except for the Algebra, this one was actually pretty easy as far as these types of problems go. Algebra's always a drag though.

Flay

Here's another one for ya...

Find the value of n for which the equation (n + 2)x^2 + 3x - 5 = 0 has one root triple the other.

KingAgamemnon

Does n have any restrictions?

Flay

Not that I'm aware of. That's the whole question.

Beedle

Flay wrote: »

Here's another one for ya...

Find the value of n for which the equation (n + 2)x^2 + 3x - 5 = 0 has one root triple the other.

I can do this (I think), though the answer is almost wholly unedifying.

First up, that equation would be more useful in the form x² + px + q = 0, so divide through by n+2. The constant term, which is now -5/(n+2), must be positive because otherwise the roots would have to be one negative and one positive (the constant term is the product of the roots). So (n +2) is negative. So at this stage I gave up on n+2, and I let m=-(n+2) so the equation was x² - (3/m)x + 5/m = 0. Calling the two roots a a and 3a, then the equation is also (x-a)(x-3a)=0. Expanding out and comparing coefficients gives 4a = 3/m and 3a² = 5/m. The first of these gives a = 3/(4m), subbed into the second gives 27/(16m²) = 5/m. Cross-multiplying and cancelling an m gives 27 = 80m, leaving us with m=27/80. Then n+2 = -27/80, and n=-187/80. Perhaps.

KingAgamemnon

Hmmmm, I want to agree with you Beedle, but (since n can be any real number) what if we set n = -2? Then when we divide by n+2 we are dividing by zero, which dear god no. Of course, if n were -2 then the 2nd degree would cancel out of the equation leaving a linear behind.

Anyway, I can't tell if you're right or not. And, I can't think of anything else, maybe some other brighter star can.

physi_marc

Beedle is correct. However, if you're dividing by (n+2), you should explicitly mention that it stands only for n != 2.

By the way, I solved the problem using the quadratic formula:

x = [-b +/- sqrt(b^2 - 4ac)] / 2a

with ax^2 + bx + c = 0.

In your case, you get:

x = [ -3 +/- sqrt(49 + 20n) ] / (2n + 4)

Now, call the - solution x1 and the + solution x2. You're looking for n such that x2/x1 = 3 (because the + solution is larger than the - solution). Solve the equation. You get n = -187/80, as Beedle already found.

BlochWave

Well he's not asked to do anything fancy like find where the quadratic's solutions exist or anything that would conflict with the fact that at n=-2 it stops being a quadratic(and hence if you're looking for two solutions you're gonna get a kick in the balls)

So to make sure I understand, you want to find the two solutions(r1 and r2) to this equation, and you want n such that r1=3*r2(or I guess r2=3*r1)?

So just solve it, you can directly apply the quadratic equation if you wish(though you divide by 2(n+2) leading to the same restriction on n that, as mentioned, we don't particularly care about)

{-3+/- sqrt[9-20*(n+2)]}/[2(n+2)]

You could divide by n+2 and get rid of the a term for the first step if it makes you feel better, had I done it I would've had half a dozen other slashes to type

So there are your two solutions(varying by the + and -)so set one equal to three times the other and solve for n. If that turns out to be impossible, set the other equal to 3 * the other and solve for n

EDIT: Well the other guy beat me but I'm leaving this so I can be a poor sport

Flay

Sorry guys, I think I've been taught a different method of solving these types of problems. Thankfully I've got a solution now, as follows;

x^2 + (3/(n+2))x - 5/(n+2) = 0

a+3a = -3/(n+2)
a*3a = -5/(n+2)

4a = -3/(n+2)
3a^2 = -5/(n+2)

Simultaneous equation, so isolate 1/(n+2):
4a/-3 = 1/(n+2)
3a^2/-5 = 1/(n+2)
So,
4a/-3 = 3a^2/-5
20a=9a^2
9a^2-20a=0
a(9a-20)=0
a=0 or a= 20/9
Sub into one of the initial equations to get:
Inconsistency, and:
-80/27=1/(n+2)
n=-197/80

Sub this back in: -(27/80)x^2 + 3x - 5 = 0 and solve to get roots of x= 2.22 and 6.66

Thanks for your help, though.

Flay

Damn, stuck again. Here's the problem;

Solve 2^(2x +1) - 5 x 2^x + 2 = 0

I can see it's an equation reducible to a quadratic, the problem is the form it's in...

Beedle

Is the x between the 5 and the 2 a letter x or a times sign? If it's a times sign, that would seem to make it significantly easier.

BlochWave

First you should get everything of the form 2^something(assuming that x is a multiplication sign, which I guess it is from your formatting)

Clearly 2=2^1, that was easy, two of them are already in the proper form, what about the 5? Remember 5=2^2+1, and multiply that by 2^x, and remember that 2^x*2^y=2^(x+y)


Is it more apparent after that?

Beedle

Assuming it is a times sign still (which there is a proper symbol for: ×), there's no need to put everything as a power of 2 (I don't see how that would help at all in fact). Just make a substitution y=2^x, and you get a quadratic in y that you can solve (then remember to put it back in terms of x).

Flay

Whoops, I forgot to reply before, but I did solve that last problem; thanks for your help. It's pretty easy when you think about it, I was just being lazy.

Now, however, I have yet another problem. Here's the question;

Find the derivative of 3 cos^3 5x

And that's not cos to the power of 35x, if anyone was wondering.

TechBoy

3 cos^3 5x is the same thing as 3 (cos 5x)^3

Whip it good with the Chain Rule.

Flay

Got it, thank ye kindly good sir.

However, I seem to have run in to more complications;

Find the equation of the tangent to the curve y = sin(pi - x) at the point (pi/6, 1/2), in exact form.

I'm fairly familiar with this type of question, but I'm a little shaky on exactly how to differentiate equation, and exactly how the result will fit in to the y - y1 = m(x - x1) format.

Beedle

Well, you can take the derivative just using the chain rule (remember, pi is just a number, treat it as if it were a 3 written there). Now, the value of the derivative at a particular x is by definition the slope of the tangent at that point, so the value of your derivative function at x=pi/6 gives you your m. And your y1 and x1 are given to you, so you're finished.

Flay

Sorry about the late reply, but thanks again.

And here we go, one more time;

A store offers a hire-purchase loan on a lounge suite costing $5000. If you pay $200 deposit, the loan on the balance is paid over 3 years, at 18.5% p.a. interest. You do not have to make a payment for the first two months. What are the monthly repayments, and how much money will you pay for the lounge suite?

I just can't figure this bloody thing out, my calculations keep turning out all wrong. The answer for the monthly fees should be $188.12, but for some reason I keep ending up with $2600...

Flay

Never mind the last question, I've gots me a new problem;

Find the greatest and least values of the function f(x) = 4x^3 - 3x^2 - 18x in the domain -2 ≤ x ≤ 3.

EDIT: Man, I'm bad at this subject. Two more questions;

Find the maximum possible area if a straight 8m length of fencing is placed across a corner to enclose a triangular space.

... and...

Find any stationary points or inflexions on the curve y = x(x - 2)3

For this last one I just end up with stat points at x = 0 and 2, where they should be points of inflexion at x = 2 and 1, and a minimum at 1/2.

HalberdBlue

Do you know what critical points are and how to find them? That will solve the first and second one for you.

more hints on the second one:

The equation is the area of a triangle, and you need to write either the base or the height in terms of the height or the base respectively. You can do this since you know the length of the hypotenuse is constant.

I don't think I ever learned about what the 3rd question is asking.

urahonky

Ugh. This sounds alot like Calculus II that I just went through. If you've not checked out this site:

www.cramster.com

I highly recommend it. If these questions you've posted are in a text book, chances are they are on that site. 9 times out of 10 they walk you through the problem so that you can see what's going on.

Flay

I really need some help with this question...

Find the volume of the solid formed when the curve y = (x + 5)^2 is rotated about the y-axis y = 1 to y = 4

Show your working, if you could.

FunkyWaltDogg

Someone stop me if I'm wrong, but I believe that you will need to solve for x since you are rotating around the y-axis.

y = (x + 5)^2
x = (+/-)sqrt(y) - 5

Here's where things get a little hazy. You only need to keep one branch; I think you will keep the negative one since it "stands out" from the axis more, so the solid you make with that one will enclose the solid from the positive branch. Hopefully you can get it from there.

Flay

Another question;

Find the derivative of x√(x + 3)

It should be simple, but I can't seem to get the answer: 3(x + 2)/2√(x + 3)

EDIT: Never mind, got it.

Flay

Having some trouble with this question;

Find all points of inflexion on the curve y = 3cos(2x + pi/4) in the region 0 ≤ x ≤ 2pi.

Show your working, please.

FunkyWaltDogg

Take the derivative and set it equal to zero, of course!

dy/dx = -3*sin(2x + pi/4) * 2 [chain rule] = -6*sin(2x + pi/4)

-6*sin(2x + pi/4) = 0
sin(2x + pi/4) = 0

sin(u) = 0 when u = -pi, 0, pi, 2pi, etc., so just solve for x using various u and keep the ones in the domain.

For example, u = pi, 2x + pi/4 = pi => x = 3pi/8

RocketScience

FunkyWaltDogg wrote: »

Take the derivative and set it equal to zero, of course!

dy/dx = -3*sin(2x + pi/4) * 2 [chain rule] = -6*sin(2x + pi/4)

-6*sin(2x + pi/4) = 0
sin(2x + pi/4) = 0

sin(u) = 0 when u = -pi, 0, pi, 2pi, etc., so just solve for x using various u and keep the ones in the domain.

For example, u = pi, 2x + pi/4 = pi => x = 3pi/8

At a point of inflexion the 2nd derivative is also equal to zero.

You do have a maths textbook, don't you Flay? This is the kind of stuff you'll find if you read one.

FunkyWaltDogg

RocketScience wrote: »

FunkyWaltDogg wrote: »

Take the derivative and set it equal to zero, of course!

dy/dx = -3*sin(2x + pi/4) * 2 [chain rule] = -6*sin(2x + pi/4)

-6*sin(2x + pi/4) = 0
sin(2x + pi/4) = 0

sin(u) = 0 when u = -pi, 0, pi, 2pi, etc., so just solve for x using various u and keep the ones in the domain.

For example, u = pi, 2x + pi/4 = pi => x = 3pi/8

At a point of inflexion the 2nd derivative is also equal to zero.

You do have a maths textbook, don't you Flay? This is the kind of stuff you'll find if you read one.

My bad, I haven't done this stuff in ten years. Just take the second derivative then, it's no more difficult.

d2y/dx2 = -12cos(2x + pi/4)

cos(u) = 0 at pi/2, 3pi/2, 5pi/2, etc.

u = pi/2, 2x + pi/4 = pi/2 => x = pi/8, etc.

FunkyWaltDogg

Hmm, there appears to be more to it still. Check it out at Wolfram Mathworld.

In fact, bookmark that site and check it for anything, it is awesome.