Need Math Help?

[Death]

Ahhhh, thanks a lot man, appreciate it. If you wouldn't mind helping me with a couple more (if you can ):

Bob and Bill both swim at 0.50 km/h. They start beside a river 015 km wider with a current of 0.30 km/h. Bob swims 150 m directly downstream, then turns around and swims back to his original spot.

Bill swims to a tree directly across the river, then turns around and swims back to the original spot. How long does each person take to complete their trips? (0.94 h, and 0.75 h)

Answers are given but they seem so random. For example, Bill is swimming straight across which is a distance of 0.15 km, and his speed is given. How come it doesn't work to find the time it takes him to swim across (delta d / v) and multiply by 2 to account for his return trip??

A boy throws newspaper sideways onto porches of customer while riding his bike, which is 15 m in front of the porches. The boy throws the papers at a horizontal speed of 6.0 m/s relative to himself and rides the bike at a speed of 4.0 m/s relative to the sidewalk.

How far in advance of a porch should he throw a paper so it lands on target? (10 m)

If he waits until he is directly opposite a porch, at what horizontal angle with respect to the sidewalk will he have to throw the paper to hit the porch? (48.2 degrees)

I have no idea how to solve this one. Should I first find the speed with which the papers travel relative to the ground? And will this even help?????

 

[AnDaLe]

wow this thread is awesome, i'm gonna start asking for nonsense calculus homeowrk on here now.

EDIT: jsut foudn out about it now

 

[Super Touhey]

Ok, Bob first swims with the current causing him to go at (.5+.3) km/h, or v1 = .8 km/h. When he goes upstream, he goes against the current at (.5-.3) km/h, or v2 = .2 km/h. Distance, d, equals .15 km.

t1 = d/v1 = .15/.8 hrs = .1875 hrs
t2 = d/v2 = .15/.2 hrs = .75 hrs
Bob's total time is t1 + t2 = .9375 hours.

What you did wrong in Bill's problem is assume he is swimming straight across the river. Bill is actually swimming upstream slightly so that his Vy cancels out with the current. Vy has to be .3 km/h. Since his total velocity is .5 km/h, Vx must be .4 km/h, based on the simple 3,4,5 right triangle. He does this going both ways so you can just multiply by two.

t = 2*d/Vx = .3/.4 hrs = .75 hours.

Make sense now? If you need help understanding it, I suggest asking the teacher or a fellow student to explain vectors to you more clearly. It really makes these problems much easier when you understand them.

For the newspaper problem, the first part is really easy. When he throws the paper, the velocity of the bike is also applied to the velocity of the paper. It travels in the x direction at 6 m/s, and the y at 4 m/s. Dx/Vx = Dy/Vy. Dy = Vy*Dx/Vx = 4*15/6 m = 10 m.

The other part is easy once you know what they're talking about. He has to throw the paper backwards at an angle so that the applied 4 m/s is canceled out. When you draw that triangle, you see that you have to take the arc cosine of 4/6. Arccos(2/3) = 48.19 degrees.

 

[Corpsecreate]

"Your road train carrying 10,000 kg of sugar (and your pet mule) breaks down 1,000 km from
your destination. Your RAC membership has lapsed, so you decide to use the mule to
transport as much of the sugar as you can the rest of the way. Fortunately the road is level and
dead straight, but unfortunately the mule can’t carry more than 1,000 kg at any given time,
and it will continuously eat the sugar it’s carrying, at the rate of 1 kg per kilometre travelled.
Also, your pet will mulishly refuse to move in any direction at all if it has no sugar to eat.
Assuming that the only means of transporting the sugar is using the mule (you can’t carry
any, the mule can’t pull the road train, and there’s nobody else to help) and that you cannot
prevent the mule from eating the sugar, what is the maximum amount of sugar, if any, that
you can deliver to your intended destination?"

I get very close to 1400 kg. Pretty sure its right, would like to see what you guys might get for it though.

 

[AltF4]

"Your road train carrying 10,000 kg of sugar (and your pet mule) breaks down 1,000 km from
your destination. Your RAC membership has lapsed, so you decide to use the mule to
transport as much of the sugar as you can the rest of the way. Fortunately the road is level and
dead straight, but unfortunately the mule can’t carry more than 1,000 kg at any given time,
and it will continuously eat the sugar it’s carrying, at the rate of 1 kg per kilometre travelled.
Also, your pet will mulishly refuse to move in any direction at all if it has no sugar to eat.
Assuming that the only means of transporting the sugar is using the mule (you can’t carry
any, the mule can’t pull the road train, and there’s nobody else to help) and that you cannot
prevent the mule from eating the sugar, what is the maximum amount of sugar, if any, that
you can deliver to your intended destination?"

I get very close to 1400 kg. Pretty sure its right, would like to see what you guys might get for it though.

Perhaps I'm reading the problem wrong, but how could you transport any of the sugar? The mule only carries 1000 kg at a time. It has to go 1000 km, consuming sugar at a rate of 1 kg / km. So after 1000 km, he'll have consumed the entire 1000 kg of sugar he was carrying. At which point the mule will stubbornly refuse to move.

Is there a units error in the problem? Maybe it meant to say the distance is 1000m (not 1000 km)? Or am I reading into this incorrectly?

 

[Super Touhey]

You have to make small trips to get the most sugar back. Actually the smaller the trip, the more sugar you retain. I wrote the following code (in drscheme):

(define (f sugar distance)
(cond
[(>= sugar (+ (* 2 distance) 1000))
(+ (f (- sugar 1000) distance)
(- 1000 (* 2 distance)))]
[else (- (min 1000 sugar) distance)]))

(define (f2 sugar total distance)
(cond
[(>= total distance)
(f2 (f sugar distance) (- total distance) distance)]
[else (f sugar total)]))

(define (find-total-sugar distance) (f2 10000 1000 distance))

The f function takes how much sugar you can take from the given amount of sugar going the stated distance. The f2 function takes how much sugar is left, the total distance to the destination, and the distance you plan to go each time you move the sugar, and returns how much you have when you reach the end. The find-total-sugar function takes the distance you plan to make each trip and finds out how much sugar you end up with at the end.

The closer distance got to zero, the closer it got to 1400 kg.
For distance = 1 km, I got 1398 kg of sugar. For distance = .01 km, I got 1399.74 kg. So I'm getting the same answer as you are. Unless it's better to drop sugar at different places, which I highly doubt, that should be the right answer.

 

[Corpsecreate]

I solved it quite a bit differently but I understand why your method gives the same answer as what I got.

 

[Super Touhey]

I love how there can be lots of different ways to solve the same problem. Out of curiosity, how did you get your answer?

 

[Corpsecreate]

I love how there can be lots of different ways to solve the same problem. Out of curiosity, how did you get your answer?

I did it this way. The maximum that you can carry at any one time is 1000kg and so whenever you want to transport the suger to a location it is always optimal to carry 1000kg. I noticed that if you move the suger 1km in front of the start then you have to cross the 1km 19 times (because you have to carry 1k there, another 1k back, again 1km there, another 1km back and you do this 19 times because the last trip across the 1km you dont need to return).

Since the amount of suger you lose is equal to the distance you travel then that means that you will have a pile of suger of 10000-19*1km = 9981kg. The problem is that if you now move 1km then the final trip you can only take 981kg with you which is not optimal. To make each trip optimal you want your next trip to hold a multiple of 1000. So you start with 10000 and you want 9000 at your next pile and you have to make 19 trips across a distance x. 10000-19X = 9000 ---> X = 1000/19

Now for the second trip you only need to make 17 trips because you have less suger to transport and as last time, you want the next pile to be another multiple of 1000. 9000-17X = 8000 ----> X=1000/17. So the total distance I have travelled up until this point in time is 1000/19 + 1000/17. If you continue this pattern and you get common denominators then you get total distance travelled = 11638496000/14549535 ~ 800km, distance remaining to get to your goal is 2911039999/14549535 ~ 200km

By this point you have 2000kg of suger left and since now this is your final trip, you no longer need to make sure that the next pile is a multiple of 1000 and so the amount you can get to your goal where you make 3 trips is 2000-3*(2911039000/14549535) = 1357730200/969969 = 1399.77kg

 

[AltF4]

Curious. So there is no solution to the problem then. It asked you yo maximize the amount of sugar, but you can't.

 

[Oracle]

This isn't technically math, but for bonus points in my precal class we need to make a math themed pumpkin. Any ideas?
My girlfriend suggested I just carve pi into it and laugh at my awful pun. I'm thinking about somehow carving a mathematical fallacy
1*0=2*0
(1*0)/0=(2*0)/0
Zeros cancel
1=2

 

[Corpsecreate]

Curious. So there is no solution to the problem then. It asked you yo maximize the amount of sugar, but you can't.

There is a solution, the maximum you can get to your destination is 1357730200/969969 which is roughly 1400kg.

 

[AltF4]

I thought you said the solution increases as you take incrementally smaller steps. So wouldn't it then only asymptotically approach a value? Unless you're imposing integer constraints.

 

[Corpsecreate]

I thought you said the solution increases as you take incrementally smaller steps. So wouldn't it then only asymptotically approach a value? Unless you're imposing integer constraints.

I didn't say that. Super Touhey did.

There are no integer constraints in this question. Touhey and I both solved it differently but arrived to the same answer.

 

[Super Touhey]

I see how it's done now. Yeah, unfortunately, my method never came to a final answer. I guess I was wrong about not dropping sugar at different distances. I knew there had to be some way to maximize it, I just didn't see it. It makes sense to make sure you always have sugar in multiples of 1000.

 

[Black Waltz]

If f'' is continuous, show that

lim x->0 (f(x+h) - 2f(x) + f(x-h))/h^2 = f''(x)

I'm not sure how to approach this.

 

[moogle]

lim h->0 [f(x+h) - 2f(x) + f(x-h)] / h^2

split apart the -2f(x) term like this
= lim h->0 {[f(x+h) - f(x)] - [(f(x) - f(x-h)]} / h^2

move one of the h's like this
= lim h->0 {[f(x+h) - f(x)] / h - [(f(x) - f(x-h)] / h} / h

use definition of derivative twice
= lim h->0 [f'(x) - f'(x-h)] / h

and once more
= f"(x)

 

[metalmonstar]

How do you explain inverse and negative exponents to someone?

How would you teach someone to factor?

 

[cutter]

How do you explain inverse and negative exponents to someone?

A negative exponent is just taking 1 over the base and then powering it like normal.

In other words:

x^-y = 1/x^y

Plug in some numbers to try it out:

10^-2 = 1/10^2 = 1/100

"Inverse exponents" are commonly called logarithms, or logs for short. A log looks something like this:

logx(z) = y

x = Base
y = exponent
z = Argument (the answer)

This is the same thing as saying: x^y = z

Logs are mainly used when you know the base and argument, but not the exponent that is needed to raise the base to your answer.

How would you teach someone to factor?

I would suggest looking at each term and finding the greatest common multiple that each of them share. Remember, you don't have to factor out the entire polynomial all at once; you can start bit by bit and factor out one thing at a time until it's been fully factored.

 

[Death]

So the given is 1/512x^9 - 512

How do I approach this problem? This is how I wrote it:

= ((1/2x)^3)^3 - ((2)^3)^3

Where do i go from here? Straight to substituting for (a-b)(a^2 +ab+b^2) ?

 

[Yink]

So the given is 1/512x^9 - 512

How do I approach this problem? This is how I wrote it:

= ((1/2x)^3)^3 - ((2)^3)^3

Where do i go from here? Straight to substituting for (a-b)(a^2 +ab+b^2) ?

Well...your wording is a bit confusing for me, what is it equal to?

If it's equal to 0 then x = either undefined / infinity ... this is because if x was anything less than infinity the problem can't happen. I don't know how your teacher wants it though. But it makes sense because if even if x = 10,000 it'd still be too small a number.

If it's equal to 1 ... It's not too hard, I'd do it like this ->

1 = 1/512x^9 - 512 (multiply both sides by 1, getting rid of the 1 on the right)

1 = 512x^9 - 512 (add 512 to both sides)

513 = 512x^9 (divide by 512 to both sides)

1.002 = x^9 (root to the power of 9 both sides)

8.961 = x

...if that's not how it was supposed to be done, I'm really sorry I couldn't be of help.

 

[Corpsecreate]

Well...your wording is a bit confusing for me, what is it equal to?

If it's equal to 0 then x = either undefined / infinity ... this is because if x was anything less than infinity the problem can't happen. I don't know how your teacher wants it though. But it makes sense because if even if x = 10,000 it'd still be too small a number.

If it's equal to 1 ... It's not too hard, I'd do it like this ->

1 = 1/512x^9 - 512 (multiply both sides by 1, getting rid of the 1 on the right)

1 = 512x^9 - 512 (add 512 to both sides)

513 = 512x^9 (divide by 512 to both sides)

If it were equal to 1 you would get: 1 = 1/512x^9 - 512

mutiply by 512 to get: 512 = x^9 - 512^2

add 512^2 to both sides gives: 512^2 + 512 = x^9

Take the 9th root of the LHS to get x; x = ninth root (512^2 + 512)

x= 4.000867303

1.002 = x^9 (root to the power of 9 both sides)

8.961 = x

...if that's not how it was supposed to be done, I'm really sorry I couldn't be of help.

Sorry man but your working has many holes in it.

1/infinity gives 0 so saying that 1/infinity - 512 = 0 is silly, you would get -512.

multiplying everything by 1 does nothing, it doesn't get rid of the numerator of 1 at all, that is a very incorrect operation there.

As for the question itself, as you said, what is it equal to?

If it was equal to 1 then you get : 1 = 1/(512x^9) - 512

Multiply both sides by 512 to get: 512 = 1/x^9 - 512^2

Add 512^2 to both sides gives you: 1/x^9 = 512^2 + 512

Take the ninth root of everything to get x: x = Ninth root [ 1 / (512^2 + 512) ]

x = 0.2499

if it were equal to zero then that you have: 0 = 1/(512x^9) - 512

In other words you get: 1/(512x^9) = 512

Multiply both sides by 512 to get: 1/x^9 = 512^2

Take the ninth root to get: x = Ninth Root [ 1 / 512^2 ]

If you want to solve that without a calculator then its the same as writing: x = Ninth Root (1) / Ninth Root (2^18) = 1 / 2^2 = 1/4

x = 1/4

 

[Corpsecreate]

EDIT: Unintentional Double Post.

 

[Yink]

Forgive the fact that I haven't done math in a couple years.

And as for the infinity / undefined, I understood it as 1/512x^9 - 512 is the WHOLE part. So basically I thought he said (if it were equal to 0):

0 = 1/512x^9 - 512 ... where 512 was part of the denominator. That is very much so my fault I apologize. Still yeah I don't know what he put it equal to.

 

[Dark_Bomb]

Oh heck yeah! I love this topic. I myself am a second year student at Berkley University in the biology program with a double minor in mathematics and physics. I'll also help out along with the others people who help. I've never seen a topic like this before. I'll be sure to keep active here.

My specialties -
Algebra
Calculus
Statistics and Probobility
Trigonometry
Geometry
Arithmatic.

 

[#HBC | Mac]

5. Solve the recurrence T(n) = 2T(n/2) + n2.
6. Solve the recurrence T(n) = T(n − 1) + 1.

missed a lecture and I now have no idea how to solve recurrences... help?

 

[L__]

For each equation, state the number of complex roots, the possible number of real roots, and the possible rational roots.

x^3 + 4x^2 + 5x - 1 = 0

I would want to help so much, but I'm most likely an idiot compared to the math whizzes here.

 

[Death]

This is just a random question I made up. I know y = x^x can be solved but how about this?

Will it not just continue forever without ever getting any numbers out of it? Because, if we sub in "2" for x, we get y = 2^y and if that is true, then y = 2^(x^y) and on and on and on....

 

[Corpsecreate]

No it wont because if y = x^x and x = 2 then y = 2^y is incorrect and it should be y = 2^rooty. which will give 4.

How about this question? Limit as x--> infinty of (1+1/x)^x

substituting x = infinity yields (1+0)^infinity = 1 but 1 is not the answer

 

[NintendoMan07]

Ok, so my physics prof puts up homework online and the students get to have fun putting up with an extremely unforgiving system that may or may not take our answers, and when we get it wrong, there's absolutely NO explanation. At least we get 20 tries to balance things out a bit, but... it wasn't too helpful with this one that I just officially got WRONG.

"When we eventually establish lunar colonies, people living there will need to have their blood pressure taken. Assume that we continue to express the systemic blood pressure as we now do on earth (see previous problem) and that the density of blood does not change. Suppose also that normal blood pressure on the moon is still 120/80 (which may not actually be true).

If a lunar colonizer has her blood pressure taken at her upper arm when it is raised 25.0 cm above her heart, what will be her systemic blood pressure reading, if she has normal blood pressure? The acceleration due to gravity on the moon is 1.67 m/s^2."

And it asks for the answer as systolic then diastolic pressure in torr.

Oh, and the amusing thing is... well here's the "previous problem":

"A hospital patient is being fed intravenously with a liquid of density 1035 kg/m^3. (See the figure .) The container of liquid is raised 1.20 m above the patient's arm where the fluid enters his veins.

What is the pressure this fluid exerts on his veins, expressed in millimeters of mercury?"

Yeah... I don't see anything about taking systemic blood pressure.

tl;dr: How do you work the first problem?

 

[2.72]

No it wont because if y = x^x and x = 2 then y = 2^y is incorrect and it should be y = 2^rooty. which will give 4.

How about this question? Limit as x--> infinty of (1+1/x)^x

substituting x = infinity yields (1+0)^infinity = 1 but 1 is not the answer

This is a good example of why the "plug in infinity" approach doesn't work. I would start by taking the log, then substituting in y=1/x. The answer I get is log(your limit) = 1, so your limit is e.

 

[Frown]

When calculating the volume of solid of revolution, I need to turn a f(x) function into a f(y) function. How?

 

[Corpsecreate]

Not completely sure of what you mean so I'll solve an example question and hopefully it will help ya:

Calculate the volume of revolution of f(x) = x² from x = 2 to x = 5

Volume of Revolution about X-Axis = Int from A to B of Pi * [ f(x) ] ²

So you get the integral from 5 to 2 of Pi*(x²)² = (Pi*x^5)/5 evaluated between x = 5 and x = 2.
= 3125Pi/5 - 32Pi/5 = 3093Pi / 5

EDIT: This calculation is if your rotating the function about the X-Axis.

If your rotating around the Y-Axis then you need to change y = x² into a function of y, namely x = rooty which I just realised is what you were asking

You'd then solve the integral of Pi * [ f(y) ]² which is in this case is the integral of Pi * x.


Hows this question!?

If you have a circle of Radius 2 and centre (4,3) then what is the smallest angle of a line needed to only JUST touch the circle?

 

[M.K]

To find the sum and product of roots in a Ax^2 + Bx + C equation, I know that

Sum = -b/a

Product = c/a

Does this still apply to cubic/quartic functions? And if not, how do I find the sum/product of the roots of cubic/quartic functions?

 

[moogle]

To find the sum and product of roots in a Ax^2 + Bx + C equation, I know that

Sum = -b/a

Product = c/a

Does this still apply to cubic/quartic functions? And if not, how do I find the sum/product of the roots of cubic/quartic functions?

For a cubic equation, think of it in factored form:
Ax^3 + Bx^2 + Cx + D = k(x - r)(x - s)(x - t)
where r,s,t are the roots.

If you multiply out the factored form, you get
k(x^3 - x^2*r - x^2*s - x^2*t + xrs + xrt + xst - rst)

Group together the x^2 terms, x terms, etc.
(k)x^3 + (-kr-ks-kt)x^2 + (krs+krt+kst)x + (-krst)

Set this equal to the original Ax^3 + Bx^2 + Cx + D, and you get
A=k
B=-kr-ks-kt
C=krs+krt+kst
D=-krst

Now use these pieces to find what you want to know.

To find the sum of the roots, r+s+t, do
-B/A = -(-kr-ks-kt)/k = r+s+t

To find the product of the roots, rst, do
-D/A = -(-krst)/k = rst

So for a cubic equation, sum=-B/A, product=-D/A. For a quartic, sum=-B/A, product=E/A. For a quintic, sum=-B/A, product=-F/A. See the pattern? (Note that the sign alternates on the product formula.)


p.s. You can do also fun stuff like finding the sum of the reciprocals of the roots.
Like for the cubic, -C/D = -(krs+krt+kst)/(-krst) = 1/t + 1/s + 1/r

 

[starmatrix]

Can someone explain to me when exactly a function has an inflection point, and how to tell mathematically?

 

[Corpsecreate]

An inflection point is when the gradient of the gradient is zero. Meaning that the rate at which the gradient curve of the original function is 0. It's a little weird to explain, its the point where the gradient moves from positive to negative or from negative to positive. You can see this on a graph by looking at the point where the curve 'changes direction'.

Like in this picture it seems as though the graph turns when x = 0.

To solve exactly where the inflection point is we find the 2nd derivative.

y = x^3 - 6x - 4
y' = 3x^2 - 6
y'' = 6x

Set the 2nd derivative (y'') to zero and solve for x.

--> 0 = 6x
x = 0

So the point of inflection occurs when x = 0 and by substituting x = 0 into y we get y = -4

The point of inflection is then (0, -4)

 

[AnDaLe]

The impedance of a circuit is given by the formula Z = √(R² + X²) where Z, R, and X are the impedance, resistance, and reactance respectively.

Resistance and reactance are both changing relative to time. Find a formula relating the rates of change of impedance, resistance, and reactance.

 

[`Jammin' Jobus]

can someone gimme a quick run down of how to change the order of double and triple integrals.

I missed a lecture in vector calc and am a little lost.

From what i gather you need to draw and picture and just sort of think of a new domains to integrate your elementary region over in a new order. anyone got any tips? there is not a magical formula.

 

[LAX_]

i like to think i am good at math (have always been in the 95% range in all my honors classes blah blah) and my first test in bc calc was a 95 (simple test on derivatives) but i BOMBED my integral test. got a 60%, and a warning sent home. it seriously freaked me out. calc is so different than all the other math areas.