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[FortuneFaded]

Probably missing something absurdly simple. I don't understand what he did here to simplify the equation. Thanks in advance

1−​2/ (√​(x+1))​​​​​​​=0

√​(x+1) ​​​=2

x=3

The step between the 1st and 2nd was taking the 2/ (√​(x+1))​​​​​​​ to the other side so that:

2/ (√​(x+1))​​​​​​​ = 1

Then he times both sides by (√​(x+1)) (assuming x =/= -1) so

√​(x+1) ​​​=2

Then squaring both sides gives:

x+1 = 4

 

[Newt]

Edit: beat

 

[Permanently A]

The step between the 1st and 2nd was taking the 2/ (√​(x+1))​​​​​​​ to the other side so that:

2/ (√​(x+1))​​​​​​​ = 1

Then he times both sides by (√​(x+1)) (assuming x =/= -1) so

√​(x+1) ​​​=2

Then squaring both sides gives:

x+1 = 4

Oh man I'm dumb. I multiplied by √​(x+1) but forgot to cancel it... this is what happens when you study for tests on the day of

don't be like me kids, get that studying done well beforehand

thanks again

 

[triplestation]

nvm

 

[Averon]

.

The box is where I got up to and the purple is the solutions manual solution for completing the problem which i dont understand.

f(1) = 10 is the value of the function you calculated in the box at x=1. That is, f(1) = 4(1)^(3/2) + 2(1)^(5/2) + C = 4 + 2 + C.

Since we know f(1) = 10, we have 10 = 4+2+C. Solving for C, we get 10-6 = C = 4.

Since C = 4, we get 4x^(3/2) + 2x^(5/2) + 4.

 

[triplestation]

f(1) = 10 is the value of the function you calculated in the box at x=1. That is, f(1) = 4(1)^(3/2) + 2(1)^(5/2) + C = 4 + 2 + C.

Since e know f(1) = 10, we have 10 = 4+2+C. Solving for C, we get 10-6 = C = 4.

Since C = 4, we get 4x^(3/2) + 2x^(5/2) + 4.

I just realized (1)^3/2 is the square root of 1^3 and (1)^5/2 is the square root of of 1^5

-__-

Still though, thanks a bunch

 

[Earthstrike]

nvm

So The function you've integrated is f(x). Since f(1)=10, plug one into f(x) and solve for c. Substitute c into f(x) and you've solved.

 

[MathUser]

Don't let my name fool you. I'm actually not that proficient with math.

 

[VegiHam]

So how do I prove that bn/an tends to 0 and an + bn tends to infinity if an tends to infinity and bn tends to a real number as n tends to infinity?

Best I can think of is writing "because common sense duh" and leaving it there...

 

[Therion]

So how do I prove that bn/an tends to 0 and an + bn tends to infinity if an tends to infinity and bn tends to a real number as n tends to infinity?

Best I can think of is writing "because common sense duh" and leaving it there...

Suppose b_n tends to B. Let E>0. Then there exists N_1>0 such that |b_n-B|<E for all n>N_1. Since a_n tends to infinity, there exists N_2>0 so that a_n>(|B|+E)/E for all n>N_2. Let N=max{N_1,N_2} and look at |b_n/a_n| when n>N.

The other question should be easier.

 

[VegiHam]

Suppose b_n tends to B. Let E>0. Then there exists N_1>0 such that |b_n-B|<E for all n>N_1. Since a_n tends to infinity, there exists N_2>0 so that a_n>(|B|+E)/E for all n>N_2. Let N=max{N_1,N_2} and look at |b_n/a_n| when n>N.

The other question should be easier.

Oooooh. I see it! Cheers!

Is the second one just algebra of limits?

 

[jnWake]

Second one is similar.

Take M>0, then there exists N_1 such that |b_n-B| < 1/M, or b_n > B-1/M. There also exists N_2 such that a_n > M - (B-1/M). Take N=max(N_1,N_2), and you have

a_n+b_n > M.

You could also try by using contradiction, but then you'd have to assume a_n+b_n converges.

 

[MisterLuffy]

I need help on this problem:

How many 16 bit strings of 0's and 1's contain exactly 7 1's?
How many 16 bit strings of 0's and 1's contain at least one 1?

I don't know what to do.

 

[Snow]

I need help on this problem:

How many 16 bit strings of 0's and 1's contain exactly 7 1's?

How many ways are there to choose 7 from 16?

How many 16 bit strings of 0's and 1's contain at least one 1?

I don't know what to do.

How many 16 bit strings are there? There is only one string with no 1, so it's the total number of possible strings minus that one without a 1.

 

[MisterLuffy]

How many ways are there to choose 7 from 16?



How many 16 bit strings are there? There is only one string with no 1, so it's the total number of possible strings minus that one without a 1.

I found an example my instructor did. It goes something like this: 16!/9!7! for the first question. I'm not sure about the second question.

 

[Gozan]

I was reading plasma physics and there's some notation that I've never encountered before:

What the hell is that arrow doing there?

And just a line, from an earlier section:

 

[Leezard]

I was reading plasma physics and there's some notation that I've never encountered before:

What the hell is that arrow doing there?

And just a line, from an earlier section:

The arrow/line is there to show you that part of the expression becomes zero. I see it often in hand-written notation, but seldom in machine-written notation.

 

[Snow]

I found an example my instructor did. It goes something like this: 16!/9!7! for the first question. I'm not sure about the second question.

Yes, 16!/(9!7!) is the way you calculate choosing 7 from 16. You can read up on combinations if you're not familiar with them. If you're not sure why they apply to this problem: imagine you have 16 marbles each with a number printed on them for a bit position. By selecting strings with seven 1s you're picking seven marbles from those 16. Which is classic combination territory.

For the second question, you need to figure out how many possible 16 bit strings there are. If this is not immediately obvious to you, try to see how many 1 bit strings there are, how many 2 bit strings, 3-bit strings etc. to build some intuition.

I.e. possible 1 bit strings:
1
0
So only 2

Two bit strings:
00
01
10
11
So 4

Three bit strings
000
001
010
011
100
101
110
111
So 8

So every extra bit doubles the number of strings, which gives 2^n possible strings for n bits. Apply that to n=16 and subtract 1 for the 0000000000000000 string that does not include a 1.

 

[More_Badass]

Anyone here with stats/behavioral stats experience? Need to decipher my psychology experiment data and it's been a few months since I used SPSS or had to remember what P values, etc. mean

 

[MisterLuffy]

Yes, 16!/(9!7!) is the way you calculate choosing 7 from 16. You can read up on combinations if you're not familiar with them. If you're not sure why they apply to this problem: imagine you have 16 marbles each with a number printed on them for a bit position. By selecting strings with seven 1s you're picking seven marbles from those 16. Which is classic combination territory.

For the second question, you need to figure out how many possible 16 bit strings there are. If this is not immediately obvious to you, try to see how many 1 bit strings there are, how many 2 bit strings, 3-bit strings etc. to build some intuition.

I.e. possible 1 bit strings:
1
0
So only 2

Two bit strings:
00
01
10
11
So 4

Three bit strings
000
001
010
011
100
101
110
111
So 8

So every extra bit doubles the number of strings, which gives 2^n possible strings for n bits. Apply that to n=16 and subtract 1 for the 0000000000000000 string that does not include a 1.

Thank you. I understand the first question from your explanation. Same with the second question based off examples. I hate these types of problems which is a reason why I lack in problem solving.

 

[Timedog]

My physic's book doesn't have an explanation for rotational inertia that I can comprehend. It seems like it's skipping some non-obvious steps or something.

It gives a riemann sum for the moment of inertia, which makes sense, but then it gives an this integral...

I = \int r^2 dm

...and my mind melts out of my ears. This is supposed to be for a continuous solid body. The solution to that integral should be mr^2 + C. The book gives a bunch of solutions for I for different shapes that do not follow this convention. Some examples...

hollow cylinder: I = 1mr^2
solid sphere: I = (2/5)mr^2
etc...

There is no constant of integration (which maybe is fine if that always = 0) and there is a seemingly random constant multiple showing up.

now I *think* I understand what that multiple is before the mr^2 part in their examples...maybe. It seems like it's some sort of weighted average for density of mass across the entire radius. So a hollow cylinder with all mass exactly at r, gets 1. And any shape with any of it's mass at points between 0 and r will have a value of less than 1 depending on how that mass is distributed for the particular shape. This logically seems to make sense, and seems to fit the formulas they give at a quick glance. Let me know if this understanding is wrong though...

But yeah, I don't get how the solution to the integral they gave isn't mr^2 + Constant.

 

[Biggest-Geek-Ever]

My physic's book doesn't have an explanation for rotational inertia that I can comprehend. It seems like it's skipping some non-obvious steps or something.

It gives a riemann sum for the moment of inertia, which makes sense, but then it gives an this integral...

I = \int r^2 dm

...and my mind melts out of my ears. This is supposed to be for a continuous solid body. The solution to that integral should be mr^2 + C. The book gives a bunch of solutions for I for different shapes that do not follow this convention. Some examples...

hollow cylinder: I = 1mr^2
solid sphere: I = (2/5)mr^2
etc...

There is no constant of integration (which maybe is fine if that always = 0) and there is a seemingly random constant multiple showing up.

now I *think* I understand what that multiple is before the mr^2 part in their examples...maybe. It seems like it's some sort of weighted average for density of mass across the entire radius. So a hollow cylinder with all mass exactly at r, gets 1. And any shape with any of it's mass at points between 0 and r will have a value of less than 1 depending on how that mass is distributed for the particular shape. This logically seems to make sense, and seems to fit the formulas they give at a quick glance. Let me know if this understanding is wrong though...

But yeah, I don't get how the solution to the integral they gave isn't mr^2 + Constant.

I'm not as much of a physics guy as a math guy, but I'll give it my best shot:

That integral is the continuous analogue of the Riemann sum. It's physical interpretation is the summing up every piece of mass that is a perpendicular distance r from the axis of rotation (typically, the axis chosen is the one that passes through the center of mass). This is not an indefinite integral, since you are integrating over the entire shape. My guess is your textbook left off the limits of integration, since those can change depending on what shape you're looking at (I've seen an m subscript for the integral used before to signify that you are integrating over some mass). An example showing this might be a derivation where you'll see dm = &#961; dV where &#961; is the density and dV is a differential volume. And then this will be equated to a product involving dR along with some other constants. Your book hopefully has an example showing this.

The instances where this integral will equal mr^2 is for a point mass and for a hoop that has "no width." For both of these, all the of mass is an equal distance r from the axis, so the r can be taken out of the integral so, so &#8747;r^2 dm = r^2 &#8747;dm = r^2 m = mr^2.

The other equations you see are derivations using calculus. I think most of the derivations are doable with the skill set you have, but a few are easier and thus best left until a multivariable calculus class. Hopefully that was a sufficiently unsatisfactory answer, lol.

Edit: Thought I'd try to expand on that last thought. Since the distance r and volume is varying for pieces of masses for most shapes, you have to jump through some hoops to get around the fact that you are doing a multivariable integral in the context of single integrals.

 

[Gotchaye]

My physic's book doesn't have an explanation for rotational inertia that I can comprehend. It seems like it's skipping some non-obvious steps or something.

It gives a riemann sum for the moment of inertia, which makes sense, but then it gives an this integral...

...

But yeah, I don't get how the solution to the integral they gave isn't mr^2 + Constant.

To add to the post just above, I think the main thing you're missing is that r is a function of m, when you want to understand this as an integral over a bunch of blobs of mass. The first mass element m1 is at some distance from the axis of rotation r1, the second is at r2, and so on. So you don't just get r^2*m evaluated at two masses - the form of the antiderivative depends on the form of r(m).

This may be easier to see if you think about the moment of inertia as an integral over r, where m is a function of r. If an object is such that it has mass m1 at a distance r1 from the origin, m2 at a distance r2, and so on, then you can get the moment of inertia by marching away from the axis of rotation, summing up m*r^2 for all the mass contained in bigger and bigger cylindrical shells.

So for a hollow cylinder, the MoI is just m*R^2 because all of the mass is at a distance r=R. You can find the MoI of a solid disk or solid cylinder by summing up the MoIs of a bunch of hollow cylinders or hoops with varying radii. For a constant-density solid cylinder of density p=m/(pi*R^2*h), the differential mass in each hollow cylinder of thickness dr is dm = p*2*pi*r*h*dr. So the integral of r^2*dm over the mass is the same as the integral of p*2*pi*h*r^3*dr over the radius, from 0 to R. Do that out and substitute in the expression for density and you get that the MoI of a solid cylinder or disk is mR^2/2. You could find the expression for the MoI of a solid sphere by thinking of it as being composed of a bunch of disks stacked on top of each other, where each disk has a different radius.

 

[Hypertrooper]

Hey guys,

I think I got the idea, but I do not know how to continue: (haha I used Latex this time or tried)
That's what is given:

So my idea is that we take another element from the intersection which is not 0.

That means if I subtract the v from each others, it should give us 0.

I have no clue how I should continue this proof. I know that 0 is the only element of the intersection, because J_1 and J_2 are disjoint, but combined they give us I. That means J_1 can only create a part of the vectors, and J_2 creates completely different ones. So that v, we assumed to exist, cannot. (I hope you get what I mean)


btw. does anyone know a good LaTex Tutorial?

 

[FortuneFaded]

That means if I subtract the v from each others, it should give us 0.

The vi's form a basis so they are linearly independent so the only linear combination which forms 0 has 0 coefficients but you have found one which has non zero coefficients.

 

[Collete]

I'm doing a real life math problem that I can't seem to do correctly.

I have a square that is 3.5 cm by 3.5 cm.
I want to draw a perfect star inside that square with congruent sides all around.
What would be the length of a side of a star?

(Also how would I try to find out this problem for future reference?)

Spoiler

I'm so not good with math...

 

[Therion]

btw. does anyone know a good LaTex Tutorial?

I started out with "The Not So Short Introduction to LaTeX," available from any good Google search. From there, as long as you are using it somewhat regularly and looking up everything that you need but don't yet know, you can become proficient fairly quickly. At least that was my experience.

 

[Gotchaye]

I'm doing a real life math problem that I can't seem to do correctly.

I have a square that is 3.5 cm by 3.5 cm.
I want to draw a perfect star inside that square with congruent sides all around.
What would be the length of a side of a star?

(Also how would I try to find out this problem for future reference?)

Spoiler

I'm so not good with math...

Getting the answer is mostly about drawing out what you're looking for and using trigonometry. It's straightforward but might take you a while. Supposing you're after the sort of five-pointed star you get by connecting the vertices of a regular pentagon, and the star is oriented in the square such that one tip points up and the left and right arms are touching the sides of the square, you can use the formulas Wikipedia gives for the lengths of various parts of a pentagram. Here: http://en.wikipedia.org/wiki/Pentagram#Golden_ratio

The red line in the diagram on the right is going to be the same length as one of the square's sides. Then you can keep dividing by that golden ratio 1.618 until you've got the length of whichever segment it is you're thinking of as the "side" of the star.

There's sort of a problem here though because you can't draw a pentagram like that inside of a square oriented in the way I think you're thinking it should be such that all five points are touching the square. The star is going to need to be drawn at an angle for that. You can see this in the Wikipedia diagram because the two other long black segments are going to be the same length as the red segment, but they're not running straight up and down. So if the red segment is perpendicular to the sides of the square it's intersecting, the other black segments won't be able to reach from the top of the square to the bottom. Edit: I take it back - I think it's not possible to draw a regular pentagon inside of a square so that all five vertices touch the square.

 

[triplestation]

optimization problem here

i feel like i am off

 

[Charmicarmicat]

Does anyone have any good recommendations for audio learning of math? I know it's obviously a very visual subject, and currently I am studying in a regular way, but I had the idea of getting some stuff to listen to while I'm at work to help me further, Especially algebra and precalc.

 

[FortuneFaded]

optimization problem here

i feel like i am off

No issues. It doesn't matter what the specified volume is, the optimal can will always have height being twice the radius.

 

[triplestation]

No issues. It doesn't matter what the specified volume is, the optimal can will always have height being twice the radius.

thats a great tip should come to use in the future, thanks!

 

[Two Words]

I am having an optimization problem:

A linear line passes through the point (9,36) and makes a triangle in the first quadrant and the axises are sides of the triangle. Find the equation of the line that maximizes the area of the triangle.

So I have the equation for an area of a triangle as bh/2. I use the slope/intercept form to get Y and X to be a function of m(slope) and substitute it into the area function. I do the derivative and find it equals 0 at m = 4 and -4. The slope needs to be negative, so I only use m = -4. I test this is a minimum or maximum, and it is a minimum. The question asks for a maximum. But if you think about it, there can't be a maximum because as the line that is the hypotenuse approaches a vertical or horizontal line, the area grows infinitely, right? At this point, I have to believe the question has a typo.

 

[Biggest-Geek-Ever]

I am having an optimization problem:

A linear line passes through the point (9,36) and makes a triangle in the first quadrant and the axises are sides of the triangle. Find the equation of the line that maximizes the area of the triangle.

So I have the equation for an area of a triangle as bh/2. I use the slope/intercept form to get Y and X to be a function of m(slope) and substitute it into the area function. I do the derivative and find it equals 0 at m = 4 and -4. The slope needs to be negative, so I only use m = -4. I test this is a minimum or maximum, and it is a minimum. The question asks for a maximum. But if you think about it, there can't be a maximum because as the line that is the hypotenuse approaches a vertical or horizontal line, the area grows infinitely, right? At this point, I have to believe the question has a typo.

Yeah, unless there was some other restriction, I would guess at that being a typo. Was this question on mymathlab or in a textbook...?

Your answer for a minimum checks out with me also.

 

[xezuru]

Anyone know good resources on self learning Matrix/Linear algebra?
I want to get a good headstart on the fundamentals if not get the whole foundation on my own.

Khan Academy has helped with some stuff but seems like more of an extra help more than a core lesson.

 

[Two Words]

I am a little worried that I gave an unsatisfying answer for my discrete math final. Can somebody look at this and tell me if you think I'm not going about this the way they want? The problem is about mathematical induction

n^2 + n = an even number for any integer n > 0

This is what I did:

Base case- 1^2 + 1 = 2, even

Inductive Hypothesis- k^2 + k = even number for any integer k > 0

Proving (k+1)^2 + (k+1) = even number

Case 1- If k+1 is even, (k+1)^2 is even
even + even = even


Case 2- If k+1 is odd, (k+1)^2 is odd
odd + odd = even

Therefore, n^2 + n = even numberfor some integer n > 0

Yeah, unless there was some other restriction, I would guess at that being a typo. Was this question on mymathlab or in a textbook...?

Your answer for a minimum checks out with me also.

Review problems given for the final by the professors. At least I can stop worrying about it.

 

[Skittles]

Anyone know good resources on self learning Matrix/Linear algebra?
I want to get a good headstart on the fundamentals if not get the whole foundation on my own.

Khan Academy has helped with some stuff but seems like more of an extra help more than a core lesson.

https://www.youtube.com/user/MathDoctorBob/featured
I hear that guy is pretty good, has videos on both matrix and linear algebra

 

[Gotchaye]

I am a little worried that I gave an unsatisfying answer for my discrete math final. Can somebody look at this and tell me if you think I'm not going about this the way they want? The problem is about mathematical induction

n^2 + n = an even number for any integer n > 0

This is what I did:

Base case- 1^2 + 1 = 2, even

Inductive Hypothesis- k^2 + k = even number for any integer k > 0

Proving (k+1)^2 + (k+1) = even number

Case 1- If k+1 is even, (k+1)^2 is even
even + even = even


Case 2- If k+1 is odd, (k+1)^2 is odd
odd + odd = even

Therefore, n^2 + n = even numberfor some integer n > 0

That strikes me as a strange inductive proof. Induction isn't doing much work there. Case 1: n is even so 2 is a prime factor of n^2, so it's even, and the sum of two even numbers is even. Case 2: n is odd so 2 is not a prime factor of n^2... and so on.

I guess my feeling is that the way to show this inductively is to say, okay, 1^2 + 1 = 2 which is even, so it works for the first n > 0.

Then evaluate it for n+1: (n+1)^2 + (n+1) = n^2 + 3n + 2

Given that n^2 + n is even, the above must also be even because (n^2 + 3n + 2) - (n^2 + n) = 2n + 2, which is even for all n.

 

[Two Words]

That strikes me as a strange inductive proof. Induction isn't doing much work there. Case 1: n is even so 2 is a prime factor of n^2, so it's even, and the sum of two even numbers is even. Case 2: n is odd so 2 is not a prime factor of n^2... and so on.

I guess my feeling is that the way to show this inductively is to say, okay, 1^2 + 1 = 2 which is even, so it works for the first n > 0.

Then evaluate it for n+1: (n+1)^2 + (n+1) = n^2 + 3n + 2

Given that n^2 + n is even, the above must also be even because (n^2 + 3n + 2) - (n^2 + n) = 2n + 2, which is even for all n.

Do you say it is strange because it is wrong, a needlessly complicated way to do it, or an unsatisfying method of applying mathematic induction?

This is the part of mathematic proofs that I've struggled with. What is fair game to assume is true, because it is true, and what do I need to explicitly prove?

For example, am I expected to prove that an even number squared is even and an odd number squared is odd? These are facts, and they aren't facts that the question is asking to prove. Do I have to prove odd + odd and even + even are both even? If the answer is yes, then why do you not have to prove other parts of math used in proofs. When does the recursive series of "why?" end in mathematic induction?

 

[Gotchaye]

Do you say it is strange because it is wrong, a needlessly complicated way to do it, or an unsatisfying method of applying mathematic induction?

This is the part of mathematic proofs that I've struggled with. What is fair game to assume is true, because it is true, and what do I need to explicitly prove?

For example, am I expected to prove that an even number squared is even and an odd number squared is odd? These are facts, and they aren't facts that the question is asking to prove. Do I have to prove odd + odd and even + even are both even? If the answer is yes, then why do you not have to prove other parts of math used in proofs. When does the recursive series of "why?" end in mathematic induction?

It's usually not obvious what you're allowed to say is obvious, yeah. But here what looked weird to me is that you're not really using induction. Induction is about establishing that something is true for a base case, then showing that, if it is true for some n, it is also true for n+1. So then because it's true for your base case it's true for all n. Your proof that the statement is true for n > 1 doesn't care about your base case; it's not an inductive proof. In most contexts what you've done would basically constitute a proof, but my worry is that the grader will be specifically looking for an inductive proof.

 

[Biggest-Geek-Ever]

I am a little worried that I gave an unsatisfying answer for my discrete math final. Can somebody look at this and tell me if you think I'm not going about this the way they want? The problem is about mathematical induction

n^2 + n = an even number for any integer n > 0

This is what I did:

Base case- 1^2 + 1 = 2, even

Inductive Hypothesis- k^2 + k = even number for any integer k > 0

Proving (k+1)^2 + (k+1) = even number

Case 1- If k+1 is even, (k+1)^2 is even
even + even = even


Case 2- If k+1 is odd, (k+1)^2 is odd
odd + odd = even

Therefore, n^2 + n = even numberfor some integer n > 0

What I would do is, after establishing your induction hypothesis,

(k+1)^2 + (k+1) = k^2 + 2k + 1 + k + 1 = (k^2+k) + 2k + 2

By your induction hypothesis, k^2 + k is even and thus an integer divisible by 2, so let it equal 2A. Thus,
(k^2+k) + 2k + 2 = 2A+ 2k + 2 = 2(A + k +1) = even

 

[Two Words]

It's usually not obvious what you're allowed to say is obvious, yeah. But here what looked weird to me is that you're not really using induction. Induction is about establishing that something is true for a base case, then showing that, if it is true for some n, it is also true for n+1. So then because it's true for your base case it's true for all n. Your proof that the statement is true for n > 1 doesn't care about your base case; it's not an inductive proof. In most contexts what you've done would basically constitute a proof, but my worry is that the grader will be specifically looking for an inductive proof.

I see. Well that sucks. Hope I didn't miss too many points from that. Might be a difference between an A and a B in the class. Now I'm really kicking myself for missing an obvious counting question.

The question asked to determine how many ways is there to make a 5 card poker hand of K, Q, or J in a standard deck. The answer is C(12,5), but for some reason during the test I thought it was C(12,4)*8 because I thought I had to worry about having 5 cards of the same rank. Clearly that isn't a problem I had to concern myself with and my answer was too large by a factor of 5.

 

[triplestation]

im having trouble with everything in orange

would you just plug 1/2 into the s(t) funtion and 3 into the v(t) function?

 

[Granadier]

im having trouble with everything in orange

would you just plug 1/2 into the s(t) funtion and 3 into the v(t) function?

Yes you would.

Great handwriting by the way.

 

[Biggest-Geek-Ever]

im having trouble with everything in orange

would you just plug 1/2 into the s(t) funtion and 3 into the v(t) function?

Yep! If you're confused on why the negative answer is "thrown out", it's not really that time can't be negative. Your position function extrapolates where the ball would be at for times before the ball is thrown (t<0) and after the ball hits the ground (t>3). Obviously, it makes no sense to consider these time periods since the physical context of the problem makes them invalid.
If you were plotting, say, the position of a particle that you knew acted the same before you started timing it, then periods for t<0 would have physical sense.

 

[Granadier]

On the topic of math (calculus), I have a calc 1 final on Wednesday that will determine pass/fail of the course. I might just pass out from stress on exam day.

Any tips from veteran calculus takers outside of the general "study the material" or "remember the C"?

 

[Biggest-Geek-Ever]

On the topic of math (calculus), I have a calc 1 final on Wednesday that will determine pass/fail of the course. I might just pass out from stress on exam day.

Any tips from veteran calculus takers outside of the general "study the material" or "remember the C"?

If you find watching examples helpful, then I would highly recommend PatrickJMT. He's the best youtuber for math videos, IMO; very clear and in general excellent at explaining step by step how to solve a problem. Here's a link to his Calculus 1 playlist: https://www.youtube.com/playlist?list=PL58C7BA6C14FD8F48

 

[triplestation]

Yep! If you're confused on why the negative answer is "thrown out", it's not really that time can't be negative. Your position function extrapolates where the ball would be at for times before the ball is thrown (t<0) and after the ball hits the ground (t>3). Obviously, it makes no sense to consider these time periods since the physical context of the problem makes them invalid.
If you were plotting, say, the position of a particle that you knew acted the same before you started timing it, then periods for t<0 would have physical sense.

Thank you and thank you granadier!

unfortunately i think i just flunked my second exam because the problems on it were beyond what the likes of the above problem i posted

 

[Granadier]

Thank you and thank you granadier!

unfortunately i think i just flunked my second exam because the problems on it were beyond what the likes of the above problem i posted

Are you me!?

I have a very difficult time extrapolating the base concepts beyond what is taught, so I've had this exact thing happen to me.

Spoiler

twice actually, oh god my grade

 

[triplestation]

Are you me!?

I have a very difficult time extrapolating the base concepts beyond what is taught, so I've had this exact thing happen to me.

Spoiler

twice actually, oh god my grade

it also doesnt help when the professor just zips through a hard example and expects you to follow -_-