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The Math Help Thread

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Certinty said:
Ah well I'm in the UK so things are a little different here. Going to get a full list from my teacher of all the topics on Tuesday, that should help. Not really sure what specifically as I'm studying Computing but most of my first year requires maths.

Study plan really. Got 2 months to learn all of it so thought I might as well start now before it's too late.
Click to expand...

Computing is normally applied maths, and also consider the programming threads as a great resource for topic-specific relations to the maths. Here's a paper on Computational Discovery in Pure Mathematics (Queen's College), I find it helpful to get a wide view before diving into an array of specifics, to avoid: "losing sight of the forest staring at one tree."
 
Partial Gamification said:
Computing is normally applied maths, and also consider the programming threads as a great resource for topic-specific relations to the maths. Here's a paper on Computational Discovery in Pure Mathematics (Queen's College), I find it helpful to get a wide view before diving into an array of specifics, to avoid: "losing sight of the forest staring at one tree."
Click to expand...
Thanks so much, will read through that now. Much appreciated.
 
How do I take the derivative of this

-7xsin(x)cos(x)

Since I don't know what else to do with it I'm guessing chain rule, but I don't know how to do that very well yet.
 
youngwerther said:
How do I take the derivative of this

-7xsin(x)cos(x)

Since I don't know what else to do with it I'm guessing chain rule, but I don't know how to do that very well yet.
Click to expand...
Think you can just use this identity:
sin(x)cos(x)=sin(2x)/2
to substitute and then just do product rule
d/dx[(-7/2)xsin(2x)]
(-7/2)sin(2x) + (-7/2)x(cos(2x))(2)
(-7)(xcos(2x) + sin(2x)/2)
 
youngwerther said:
How do I take the derivative of this

-7xsin(x)cos(x)

Since I don't know what else to do with it I'm guessing chain rule, but I don't know how to do that very well yet.
Click to expand...


f(x)g(x)
product rule applied:
f'(x)g(x) + f(x)g'(x)

For this problem, it appears that there are three pieces:

(-7x)(sin(x))(cos(x))


f(x)g(x)h(x)
the product rule for three pieces works like this:
f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)

D'oh, accidentally called product rule the chain rule. Fixed now
 
Ok thanks.

I'm going to need a freaking inhaler soon.

Let F(x) = f(x^5) and G(x) = (f(x))^5

You also know that
a^4 = 6
f(a) = 3
f'(a)= 4
f'(a^5) = 9

Find F'(a) and G'(a)


I really don't even know. I'm so totally lost on this stuff.
Can someone explain this to me? The next one is even more complicated so I'm hoping I can do that one on my own after I decipher this.

I'm guessing to find F'(a), I need to know what a^5 is, but I don't really understand how to get that. I see they give me what the derivative of a^5 is. Can I get it from that?
 
youngwerther said:
Ok thanks.

I'm going to need a freaking inhaler soon.

Let F(x) = f(x^5) and G(x) = (f(x))^5

You also know that
a^4 = 6
f(a) = 3
f'(a)= 4
f'(a^5) = 9

Find F'(a) and G'(a)


I really don't even know. I'm so totally lost on this stuff.
Can someone explain this to me? The next one is even more complicated so I'm hoping I can do that one on my own after I decipher this.

I'm guessing to find F'(a), I need to know what a^5 is, but I don't really understand how to get that. I see they give me what the derivative of a^5 is. Can I get it from that?
Click to expand...
You don't need to know a^5. Heavy hint: find the derivative of F(x), just in terms of x.
 
youngwerther said:
Sorry, I'm utterly lost.

So if F(x) = f(x) * x^5

and f(a) = 3

then

F'(a) = d/dx (3) * ??


I cannot grasp this.
Click to expand...

No, F(x)=/= f(x) * x^5. Unfortunately that's not a correct application of the chain rule, or really one at all since you didn't take the derivative of anything.
F(x)=f(x^5). That's what we're given. So, knowing the chain rule, think of it this way:
"say F(x)=f(g(x)) where f(x) is our given, arbitrary function and g(x)=x^5"
Then, F'(x)=f'(g(x)) * (g'(x))
We said that g(x)=x^5. Then g'(x)=5*(x^(5-1))=5(x^4). Substituting in:
F'(x)=f'(x^5) * (5(x^4))
Now what happens when you evaluate F' at a? That is, what's F'(a)? if you plug in a for x in every spot in the equation right there, you should find that everything comes together pretty elegantly.

EDIT: I'll just post the rest of the work since I have it typed up in another tab anyway:
Substitute a
F'(a)=f'(a^5)*(5(a^4))
Substitute your given values. We're told what f'(a^5) is and what a^4 is.
9*(5*6)=9*30=270.

Now how about G(x)? Well here we can use those same functions f(x) and g(x)=x^5 that we had before, the key difference being that now, G(x)=g(f(x))=(f(x))^5
Then, by the same chain rule principles as before:
G'(x)=5((f(x))^4)*(f'(x))
Substitute a.
G'(a)=5((f(a))^4)*(f'(a))
Substitute given values
G'(a)=5((3)^4)*(4)=5(81)*4=1620
 
Implicit Differentiation

sin(x^(6)y^(7))=x

F'(x)=cos(x^(6)y^(7)) (6x^(5)*y^(7)+x^(6)*7y^(6)(dy/dx))=1

1-6x^(5)y^(7)/(7x^(6)y^(6)cos(x^(6)y^(7)))

sorry its really strange but the x^(6)y^(7) is broken using the product rule with the chain rule

please help :)
 
Tater Tot said:
Implicit Differentiation

sin(x^(6)y^(7))=x

F'(x)=cos(x^(6)y^(7)) (6x^(5)*y^(7)+x^(6)*7y^(6)(dy/dx))=1

1-6x^(5)y^(7)/(7x^(6)y^(6)cos(x^(6)y^(7)))

sorry its really strange but the x^(6)y^(7) is broken using the product rule with the chain rule

please help :)
Click to expand...

From here:

cos(x^(6)y^(7)) (6x^(5)*y^(7)+x^(6)*7y^(6)(dy/dx))=1

First divide everything by the cosine term. (You can then write the right hand side as secant.) Then subtract 6x^(5)*y^(7) from both sides. Then divide everything by x^(6)*7y^(6) to isolate dy/dx.
 
It's the same as writing (3-x)^-1 so you can just differentiate normally to get
(3-x)^-2
.

I never even remember the quotient rule anyway and just do the product rule.
 
Yes but you shouldn't bring it inside the bracket straight away as you have to differentiate (3-x) by the chain rule to get another -1
 
A stone dropped into a still pond sends our a circular ripple whose radius increases at a constant rate of 2 ft/s. How rapidly is the area enclosed by the ripple increasing at the end of 11s?


I got this:

a=pi(r)^(2) A'= 2pi(r) A'=2pi(2) A'=4pi*11

A'=44pi ft^(2)/sec

is this correct?
 
thequickandthedead said:
Can someone help me with this problem?

"one root of the equation ax^2 + bx + c = 0 is five times the other. Show that 5b^2 = 36ac."
Click to expand...

Takes me back to simpler times.

Vieta's formulas:

x1 + x2 = -b/a
x1*x2 = c/a

x1 = x2/5 = x

so

6x = -b/a and 5x² = c/a

Combine them and get

5 b²/a² = 36 * c/a

5 b² = 36 a*c
 
thequickandthedead said:
Can someone help me with this problem?

"one root of the equation ax^2 + bx + c = 0 is five times the other. Show that 5b^2 = 36ac."
Click to expand...

The factored form of the equation would be something like a (x - r) * (x - 5r). I haven't tried, but if you expand that and try to eliminate r, you should find that 5b^2 = 36ac. I guess.

Tater Tot said:
A stone dropped into a still pond sends our a circular ripple whose radius increases at a constant rate of 2 ft/s. How rapidly is the area enclosed by the ripple increasing at the end of 11s?


I got this:

a=pi(r)^(2) A'= 2pi(r) A'=2pi(2) A'=4pi*11

A'=44pi ft^(2)/sec

is this correct?
Click to expand...

Area is a function of radius. A(r) = pi * r^2
Radius is a function of time. r(t) = 2t, where t is in seconds and r is in feet.

Therefore area is a function of time, and you want to know dA/dt at t=11. Hope that helps.

Edit: Wait, I'm not sure if you're asked for dA/dt or dA/dr. They might not be the same. Sorry :S
 
Tater Tot said:
A stone dropped into a still pond sends our a circular ripple whose radius increases at a constant rate of 2 ft/s. How rapidly is the area enclosed by the ripple increasing at the end of 11s?


I got this:

a=pi(r)^(2) A'= 2pi(r) A'=2pi(2) A'=4pi*11

A'=44pi ft^(2)/sec

is this correct?
Click to expand...

You have to be careful, since it tells you r is a function of time, t, then you need to replace r with 2t, and use the chain rule

A=pi(2t)^2
dA/dt = 2pi(2t)(2) We used the chain rule here
dA/dt = 8pi(t)

So for t = 11 it's
dA/dt = 88pi
And we are done.

Note, we could do what you did, and sub in r = 2t later, but you must remember to use chain rule.

A=(pi)r^2
dA/dt=2(pi)r*(dr/dt) We used chain rule here. Also, we know dr/dt is 2.
dA/dt = 2(pi)r (2) Since the question gives us a time, t=11, we need r in terms of t. recall r = 2t

dA/dt = 2(pi)(2t)(2)
dA/dt=8t(pi) and when t = 11 we get the same as above, 88pi.
 
MikeDip said:
You have to be careful, since it tells you r is a function of time, t, then you need to replace r with 2t, and use the chain rule

A=pi(2t)^2
dA/dt = 2pi(2t)(2) We used the chain rule here
dA/dt = 8pi(t)

So for t = 11 it's
dA/dt = 88pi
And we are done.

Note, we could do what you did, and sub in r = 2t later, but you must remember to use chain rule.

A=(pi)r^2
dA/dt=2(pi)r*(dr/dt) We used chain rule here. Also, we know dr/dt is 2.
dA/dt = 2(pi)r (2) Since the question gives us a time, t=11, we need r in terms of t. recall r = 2t

dA/dt = 2(pi)(2t)(2)
dA/dt=8t(pi) and when t = 11 we get the same as above, 88pi.
Click to expand...


Kazerei said:
Area is a function of radius. A(r) = pi * r^2
Radius is a function of time. r(t) = 2t, where t is in seconds and r is in feet.

Therefore area is a function of time, and you want to know dA/dt at t=11. Hope that helps.

Edit: Wait, I'm not sure if you're asked for dA/dt or dA/dr. They might not be the same. Sorry :S
Click to expand...


Thanks a lot, I do understand now. :)
 
If I have h(x) = sqrt(2+3f(x))

and I need to find h'(x), then this is a chain rule problem with three portions, right?

f(x) as innermost
then 2+3f(x)
and then x^1/2

?
 
The problem I'm running into is that the value they give me for f(1) is just a constant, so if I take the derivative of the second function, I get zero, which kills all the other values, but zero isn't the correct answer for h(1). I'm probably using the chain rule incorrectly.

How should one use it with three functions?
 
youngwerther said:
The problem I'm running into is that the value they give me for f(1) is just a constant, so if I take the derivative of the second function, I get zero, which kills all the other values, but zero isn't the correct answer for h(1). I'm probably using the chain rule incorrectly.

How should one use it with three functions?
Click to expand...

You should be taking the derivative (2+3f(x))'=3f'(x). You need to differentiate before plugging in the 1. That's true of any function; if you plug in a value and evaluate the function you will get a constant, and differentiation will always give you zero.

To solve this problem, you would first take the power rule, then apply the chain rule, giving

(1/2)(2+3f(x))^(-1/2)(2+3f(x))'=(1/2)(2+3f(x))^(-1/2)*3f'(x).

Since the second function is so simple, I guess using the chain rule on that part is overkill. Differentiating the constant 2 gives 0, and the constant 3 can be pulled out of (3(f(x))'=3(f(x))'. The chain rule would be more helpful if that part was 2+3(f(x))^2 or something similar.
 
need help with this integral

integrate ((2x-3)/((x^2)+5x+4)) dx

I tried the method of substitution but I get stuck since I can't cancel the x so everything is in terms of the variable I'm using in the method of substitution.

any ideas?
 
Error said:
need help with this integral

integrate ((2x-3)/((x^2)+5x+4)) dx

I tried the method of substitution but I get stuck since I can't cancel the x so everything is in terms of the variable I'm using in the method of substitution.

any ideas?
Click to expand...

The denominator factors to (x+1)(x+4). Can you do partial fractions?
 
Error said:
need help with this integral

integrate ((2x-3)/((x^2)+5x+4)) dx

I tried the method of substitution but I get stuck since I can't cancel the x so everything is in terms of the variable I'm using in the method of substitution.

any ideas?
Click to expand...

Therion said:
The denominator factors to (x+1)(x+4). Can you do partial fractions?
Click to expand...

There is a simple way to integrate things like that:

(2x-3)/(x² + 5x + 4) = (2x-3)/((x+1)*(x+4))

(2x-3)/((x+1)*(x+4)) = A/(x+1) + B/(x+4) where A and B are unknown numbers

2x - 3 = Ax + 4A + Bx + B

2x - 3 = (A+B)*x + 4A + B

A + B = 2
4A + B = -3 →

3A = -5; A = -5/3
B = 11/3

Therefore the initial expresison equals:

-5/(3*(x+1)) + 11/(3*(x+4))

Int (-5/(3*(x+1)) + 11/(3*(x+4))) dx = -5/3 * ln (x+1) + 11/3 * ln (x+4) + const
 
I have this question in my HW.

Expand the function f (x) = cosh (ax) where a is a real number, in a Fourier series over the interval (-pi,pi)

We haven't done cosh yet... kinda lost.
 
rayner said:
I have this question in my HW.

Expand the function f (x) = cosh (ax) where a is a real number, in a Fourier series over the interval (-pi,pi)

We haven't done cosh yet... kinda lost.
Click to expand...

cosh(x) is just 1/2(e^x + e^-x). If you know the Fourier series for e^ax and e^-ax you should just be able to combine them and multiply by 1/2. (Actually you only need the series for e^ax, since your interval is (-pi, pi) you should be able to compose it with -x to get the expansion for e^-ax.) (Although compositions won't work in general because they might transform the interval, but in this case -x doesn't transform the interval (-pi, pi) so it should work)*

*PS I'm going off of like 5 years of time lapse since I learned about Fourier series, so some of this information may not be true.
 
Hello, i am new to math (22 years old)

I never learned to appreciate the subject in the past and in high school I thought that it was a waste of time because none of it applied to my interests in the real world. The classes bored me and always left me confused. I am now in the middle of algebra and early graphing, and now I'm doing quite well and actually appreciating it. Algebra is fucking awesome. I love how everything snaps together at the end of a problem.

I just would like to know what other doers of math think about the transitions from algebra to geometry to precalculus and so on. Will I enjoy other disciplines of math if I enjoyed algebra? When I look at differential equations now I go ''wtf is that'' but I know I shouldn't be bothered by it at this stage.
 
bangladesh said:
Hello, i am new to math (22 years old)

I never learned to appreciate the subject in the past and in high school I thought that it was a waste of time because none of it applied to my interests in the real world. The classes bored me and always left me confused. I am now in the middle of algebra and early graphing, and now I'm doing quite well and actually appreciating it. Algebra is fucking awesome. I love how everything snaps together at the end of a problem.

I just would like to know what other doers of math think about the transitions from algebra to geometry to precalculus and so on. Will I enjoy other disciplines of math if I enjoyed algebra? When I look at differential equations now I go ''wtf is that'' but I know I shouldn't be bothered by it at this stage.
Click to expand...

For me my favourite math topics are Linear Algebra, Calculus, and Group theory. It depends on what part of algebra you really like, because if it's just everything coming together nicely, that happens all the time in math, so you should enjoy it.

The history of math is also a very interesting topic, you'd be surprised at just how much older civilizations knew, and how it relates to what we do today.
 
MikeDip said:
For me my favourite math topics are Linear Algebra, Calculus, and Group theory. It depends on what part of algebra you really like, because if it's just everything coming together nicely, that happens all the time in math, so you should enjoy it.

The history of math is also a very interesting topic, you'd be surprised at just how much older civilizations knew, and how it relates to what we do today.
Click to expand...

Cool, I'll be starting geometry in maybe a month or two. I have a real vague idea of how much the older world knew but I understand that a lot of stuff they did was extremely ahead of their time. I think I'll really enjoy that.
 
Does anyone have any experience with Maple? I need to plot a phase portrait of a function against it's differential with implicitplot but I don't really understand what I'm doing and the only time I get an output and not an error it doesn't look right.

edit: Doesn't matter worked it out, dude who told me to use implicit plot was lying!
 
If you want to take the derivative of a function with three products, then you would do

y' = f'gh+g'fh+h'fg

right?

Can someone show me where I'm going wrong on this one then

y=x^2sinxtanx

I did

y' = 2xsinxtanx + x^2cosxtanx + x^2 sinxsec^2x

Wolframalpha gives this

x^2sinx +x^2tanxsecx + 2xsinxtanx

What is going on?
Is it a simplification thing?

I see where they get two of their terms.
One is the same as mine, the other they wrote tan as sin/cos and canceled the cos.
I don't see where they got the x^2tanxsecx one though.
 
How do you take the derivative of something like this

(3t-1)^4(2t+1)^-3

My book has an example but I don't follow what they did in all the steps.
They say you first want to use the product rule.
So I'd get this

(3t-1)^4 -3(2t+1)^-4 + 4(3t-1)^3 (2t+1)^-3

Where would I go from here?
 
youngwerther said:
How do you take the derivative of something like this

(3t-1)^4(2t+1)^-3

My book has an example but I don't follow what they did in all the steps.
They say you first want to use the product rule.
So I'd get this

(3t-1)^4 -3(2t+1)^-4 + 4(3t-1)^3 (2t+1)^-3

Where would I go from here?
Click to expand...


Ignore the second half for now and just consider (3t-1)^4. How would take the derivative of that with respect to t?

In general when taking the chain rule you treat the entire thing a series of nested functions, so

(3t-1)^4 = f(x(t)) where f(x) = x^4 and x(t)=3t-1

derivative of (3t-1)^4 = f'(x(t))*x'(t)

Taking the product rule was the first part of the chain, the f'(x(t)) part.

Essentially,
f(x(t)) * g(y(t)) -> [f'(x(t))*x'(t)] * g(y(t)) + f(x(t)) * [g'(y(t))*y'(t)]
 
Find the slope of the tangent line to the curve xy^3 + 2y + 0.384 = 0 at the point (-2,-0.2).


I took the derivative of the equation and got
-y^3/(3xy^2 + 2)

To calculate the slope at that specific point, do I just plug those coordinates into my slope?
I tried doing that and my answer was wrong.
 
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