MisterLuffy
Member
Yes. I'm going from left to right since the problem never asked me prove it from right to left.
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The Math Help Thread
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Yes. I'm going from left to right since the problem never asked me prove it from right to left.
Well, you're showing equality, which means you can work left to right, or right to left, or from both sides simultaneously, etc. I think the steps are simpler going from right to left since you don't need to recognize that those extra terms need to be inserted.
(AUB)-(BnA) = (AUB)n(B^cUA^c) = (AnB^c)U(AnA^c)U(BnB^c)U(BnA^c) = (A-B)U(B-A)
seems like a more straightforward argument to me.
The One Who Knocks
Member
(I've wrote out one of the questions and my solution above just to make it clear if anybody's confused)
Ok so we're doing Euler's Formula as part of Complex Numbers in my Linear Algebra class yet I've somewhat of an issue with successfully using it to find "roots hidden in the complex plane" (to quote my lecturer). The lecturer had stated that for |z| = 1, (z)^n = (e)^(i)(2pi/n). I'm hoping somebody could clarify why that is not necessarily the case for this one. I am almost certain it's because it's z^3 = - 8 but I'm not sure how you can rectify the formula given when it's negative. I can do it graphically perfectly fine, but I wish there was an intuitive way of doing it rather than having to draw it out every time, and I'm almost certain for more difficult problems that this could be an issue so I'm hoping to get it resolved. So, to be clear in what I'm asking, is there any way to adjust the formula (r^n)(e^2pi/n) when z = a where a < 0? Do I just subtract 2pi/n from pi?
MisterLuffy
Member
I see, I was so focused on doing the terms from left to right without even trying to go from right to left. At first I was confused on how you went from this term: (AUB)n(B^cUA^c) to this term: (AnB^c)U(AnA^c)U(BnB^c)U(BnA^c) but I understand what you did. My instructor did something similar in his example, but I never thought about it till now.
I was also confused on how (AnB^c)U(AnA^c)U(BnB^c)U(BnA^c) = (A-B)U(B-A)
But I remembered that, correct me if I'm wrong, (AnA^c) and (BnB^c) equals to empty sets. So that's leave (AnB^c)U(BnA^c) = (A-B)U(B-A)
Again, thank you for your help.
Thank you for showing me from left to right. I think it's much easier doing from right to left like Therion did.
Ugh, I feel so stuck whenever trig gets thrown into the mix. I have to prove that this is true
limit as x --> 3 of [x^3(sin(pi/x^2))] = 0
I have no idea where to begin because my trig is still pretty bad. I always feel like I don't know where to start with trig. It's not intuitively apparent to me, even if I remember most of trig.
Secondary question, I cannot remember what happens to an absolute value equation when it is divided by a negative number. I have the following inequality
|-3x + 15| < E (epsilon)
I want to algebraically set the left side of the equation to |x-5| < d (delta) by dividing both sides by -3. I ended up with d = -E/3, but I can't have delta as a negative number. What am I screwing up here?
limit as x --> 3 of [x^3(sin(pi/x^2))] = 0
I have no idea where to begin because my trig is still pretty bad. I always feel like I don't know where to start with trig. It's not intuitively apparent to me, even if I remember most of trig.
Secondary question, I cannot remember what happens to an absolute value equation when it is divided by a negative number. I have the following inequality
|-3x + 15| < E (epsilon)
I want to algebraically set the left side of the equation to |x-5| < d (delta) by dividing both sides by -3. I ended up with d = -E/3, but I can't have delta as a negative number. What am I screwing up here?
I'm not sure that is actually equal to zero.
If you want to remove the absolute value, that inequality becomesTwo Words said:
-E < -3x+15 < E
E/3 > x - 5 > -E/3, or
5 + E/3 > x > 5 - E/3
If you want to keep the absolute value, you can just do
|-3x+15|=|-3||x-5|=3|x-5|
so |x-5| < E/3 (equivalent to second line above)
Ok so I got the absolutes part right. I'll double check that other problem tomorrow to make sure there isn't a typo.
MisterLuffy
Member
I have another question: I have to show that this argument is valid, I haven't started on it yet but my instructor stated, I think, that there's two ways to answer this this question either using truth tables or using reasons in proof which is the shortcut. Can anyone help me? I do want to use the truth table but it would be a hassle since there's 5 variables and 2^5 would result in 32 rows for each function.
r => p
~w ∨ r
(t ∧ q) => (p ∨ s)
q ∨ w
~p
~r => t
∴ s
r => p
~w ∨ r
(t ∧ q) => (p ∨ s)
q ∨ w
~p
~r => t
∴ s
Hey man, where is your previous assignment? Did you figure that one out?
As for this one, I am not sure which rules of inference you are allowed to use, so this is how I would teach it to my students:
All these are premises and therefore assumed to be true:
1. r => p
2. ~w ∨ r
3. (t ∧ q) => (p ∨ s)
4. q ∨ w
5.~p
6.~r => t
Nothing except premise 5 tells us something with regards to the truth values of any atomic sentence, so by premise 5 we know that "p" is false. If "p" shows up as a consequent in a conditional, and we know it to be false, then we also know that the conditional's antecedent, here "r", is false. Therefore "~r" is true
7. ~r (By Modus tollendo tollens, from lines 1 and 5)
Now just look where "r" shows up, these are premises 2. and 6. In premise 2. we know that at least one of the disjuncts has to be true, and since "r" is false, "~w" is true.
8. ~w (By disjunctive syllogism or Modus tollendo ponens, from lines 2. and 7.)
Now for premises 6, whenever we have the antecedent of a conditional known to be true, then the consequent must also be true, here "t".
9. t (By Modus ponendo ponens, from 6 and 7.)
As "w" is false, we have a look where it appears again, and that is premise 4. Again, disjunctive syllogism, so "q" is true.
10. q (disjunctive syllogism, from 4. and 8.)
As "t" and "q" are both true, so must be "t^q".
11. t^q (Conjunction Introduction, from 9 and 10)
But that's the antecedent of premise 3, so again MPP
12. p v s (By Modus ponendo ponens, from 3. and 11.)
"p" reappears, but we know it to be false, so using the disjunctive syllogism rule again, we infer "s".
13. s (By disjunctive syllogism, from 5. and 12.)
MisterLuffy
Member
You mean the one I edited out...no but I'm receiving help from my instructor. My only problem with this one and the one I edited out was the fact that I don't know how to decipher the steps into reasons. I still don't understand the reasons you did on this one...I feel like it's going to take me awhile to understand the examples my instructor did in class and the problem you just helped me on.
Well, understanding the semantic rules of the logical operators is key. Just elaborate on which steps you don't uunderstand, if you like.
MisterLuffy
Member
Okay, lets start with 7, you started looking at the modus tollens rule
~p
r --> p
~r
Lets say p is true, then ~p is false. Then would r have to be false too?
Step 8, 10-13 are the ones I don't understand. Can I ask, does the reasons have to be in order?
Why would you say "p" is true? Premise 5 is "~p" which is simply telling us that "p" is false. Once you know if an atomic sentence like "p", "q" or "r" is true or false, you usually don't assume the other truth value. So "p", the consequent of "r->p" is false, and therefore "r" is false, too. Does that make sense?
As for the other steps: Do you know the truth tables for logical disjunction and conjunction?
MisterLuffy
Member
That makes sense. Yes. I'm just going to ask for my instructor's help on Monday. I'm still struggling understanding the problem and I'm easily getting discouraged. Sorry for wasting your time.
Edit: I did another problem and finished it. I want to check with you concordia.
p - I will take a bus
q- I will take a subway
r - I'll be late for my appointment
s - I will be broke
t - I will take a taxi
steps
1) p v q --> r
2) t --> r∧s
3) ~q ∧ ~p --> t
4) ~r
∴s
reasons
5) ~(p v q), Modus tollens of 1 and 4
6) ~7p∧~q, DeMorgan's Law of 1
7) t, MP of 3 and 6
8) ~r v s, Modus ponens of 2 and 7
9) s, Addition of 8
steps
1) p v q --> r
2) t --> r∧s
3) ~q ∧ ~p --> t
4) ~r
∴s
reasons
5) ~(p v q), Modus tollens of 1 and 4 //yes, that's good
6) ~7p∧~q, DeMorgan's Law of 1 // yep, but you are applying the law to line 5, not line 1
7) t, MP of 3 and 6 // yes. strictly speaking, you would first have to say why ~p^~q is the same as ~q^~p, but it's all good
8) ~r v s, Modus ponens of 2 and 7 //the rule is correct, but the premise 2) says "r^s"
9) s, Addition of 8 // since line 8 is not correct, this is also incorrect. What is the addition rule? Maybe you should write down what each of the operator rules mean, so I can clarify for you, if you like. The addition rule usually means you introduce the "v" sign, that is, a disjunction. But "s" is not a disjunction. Generally, it would help to know the rules you are allowed to use and their names.
1) p v q --> r
2) t --> r∧s
3) ~q ∧ ~p --> t
4) ~r
∴s
reasons
5) ~(p v q), Modus tollens of 1 and 4 //yes, that's good
6) ~7p∧~q, DeMorgan's Law of 1 // yep, but you are applying the law to line 5, not line 1
7) t, MP of 3 and 6 // yes. strictly speaking, you would first have to say why ~p^~q is the same as ~q^~p, but it's all good
8) ~r v s, Modus ponens of 2 and 7 //the rule is correct, but the premise 2) says "r^s"
9) s, Addition of 8 // since line 8 is not correct, this is also incorrect. What is the addition rule? Maybe you should write down what each of the operator rules mean, so I can clarify for you, if you like. The addition rule usually means you introduce the "v" sign, that is, a disjunction. But "s" is not a disjunction. Generally, it would help to know the rules you are allowed to use and their names.
MisterLuffy
Member
Edit: 2) t --> ~r v s
Then what rule should I use for step 8? I'm looking at the rules of inference, the rule of addition and disjunctive syllogism introduces the "v" sign.
With that edit (assuming everything else stays the same) I don't believe you can conclude s anymore.
True. Maybe he/she meant "r v s"?!
Right, I was thinking either that or ~r∧s.
r v s makes for a better problem, but ~r∧s is closer to what was originally posted.
I need some help grasping these equations. I'm taking an intro graphics course and am trying to remember linear algebra/trig from years ago. I have a camera system that works like an fps(and moves up and down, not just on the ground), so it rotates on the y and x axis. My trouble comes from creating wasd movement when I've rotated both axis'. I found these equations in a similar program and they work great, but I'm trying to figure out how they would have been derived.
if you go forward/w was pressed
x = Vsin(a)cos(b) cos(b) was indicated as a pitch factor
y = Vsin(b)
z = Vcos(b)sin(a) sin(a) was indicated as a yaw factor
a is the angle about the xaxis and b is for the yaxis. V is the scalar. Now I was able to make a 2d camera system without the y coordinates, and this one simply adds the pitch and yaw factor portions to what I've already had as well as the y component. I'm a little confused as to why this is needed, as well as what it would be derived from. I tried looking up the term pitch factor and yaw factor and nothing came up, so if there's a name for them that would be a great place for me to start.
EDIT: Alright, I looked over my work and realized the stupid mistake I made. I've figured out what my issue was. I really need to get some sleep.
if you go forward/w was pressed
x = Vsin(a)cos(b) cos(b) was indicated as a pitch factor
y = Vsin(b)
z = Vcos(b)sin(a) sin(a) was indicated as a yaw factor
a is the angle about the xaxis and b is for the yaxis. V is the scalar. Now I was able to make a 2d camera system without the y coordinates, and this one simply adds the pitch and yaw factor portions to what I've already had as well as the y component. I'm a little confused as to why this is needed, as well as what it would be derived from. I tried looking up the term pitch factor and yaw factor and nothing came up, so if there's a name for them that would be a great place for me to start.
EDIT: Alright, I looked over my work and realized the stupid mistake I made. I've figured out what my issue was. I really need to get some sleep.
MisterLuffy
Member
Hey concordia, I fully understand the problem you helped me on and I did it on my own without copying the problem you did. Me and my classmate were working on this problem together but we are stuck and I was wondering if you could check if I did this problem correctly.
Argument
H1) (~p v q) => r
H2) s v ~q
H3) ~t
H4) p => t
H5) (~p ∧ r) => ~s
∴~q
reasons
1) ~p, Modus tollens of H3 and H4
2) (~p v q) Modus ponens of H1
3) r, Modus ponens of 2 and H1
4) ~p ∧ r, conjunction of 1 and 3
5) ~s, Modus ponens of 4 and H5
6) ~q, Disjunctive syllogism of 5 and H2
Argument
H1) (~p v q) => r
H2) s v ~q
H3) ~t
H4) p => t
H5) (~p ∧ r) => ~s
∴~q
reasons
1) ~p, Modus tollens of H3 and H4
2) (~p v q) Modus ponens of H1
3) r, Modus ponens of 2 and H1
4) ~p ∧ r, conjunction of 1 and 3
5) ~s, Modus ponens of 4 and H5
6) ~q, Disjunctive syllogism of 5 and H2
Very good, except in step 2) you use the addition rule, not modus (ponendo) ponens.
MisterLuffy
Member
I've been wondering what rule I should use for that second step. So if I'm using addition rule then it would be like this: 2) ~p v q, Addition of 1
Thank you so much for, again, helping me.
Exactly. You're welcome.
boviscopophobic
Member
Is there something else that you're not including here? Assuming ABM is supposed to be an arbitrary triangle, then x could be anywhere between 1 and 7.So it's 3 am and i can't think. This problem should be basic as heck but the answer on the back of the book doesn't match mine.
I've been having trouble with 3d rotations. I made a post earlier about making a camera system, but it needs to be able to rotation and translate about its own axis compared to the world coordinate axis. So I know how to simply make one axis rotate, which would cause the other two to change, thus changing my basis, but I'm having trouble figuring out how to perform a new rotation on the new basis. So I rotate about the x axis 45 degrees. I now have a new z and y axis. Now I rotate about the y axis. I'm not sure what calculations I need to do with respect to the world coordinates to rotate on this newly made axis.
Anyone that knows how to do this or can direct me to somewhere that can help me would be great. I've been stuck on this most of the day, where I'd get the z axis to change about the x axis, but then I'd be rotating on the original y axis.
Anyone that knows how to do this or can direct me to somewhere that can help me would be great. I've been stuck on this most of the day, where I'd get the z axis to change about the x axis, but then I'd be rotating on the original y axis.
boviscopophobic
Member
You can use Rodrigues' rotation formula. Another common choice is to parameterize in terms of unit quaternions.
For your particular situation, here is a less efficient solution that you might or might not find easier to think about when getting started. The idea is to transform to the standard axes, do your rotation there, then transform back. Let i, j, and k be the standard unit basis vectors (written as column vectors) and let i', j', k' be your current rotated versions. Of course, i', j', k' should form an orthonormal basis. Define the matrix
P = [ i' j' k' ]^T
i.e., the rows of P are the transposed versions of i', j', and k'. Then P is exactly the rotation matrix that carries i',j',k' back to the standard basis i,j,k. (Why?) Now let R be the matrix of your desired rotation around j (the standard y-axis). The "conjugation of R by P", that is, P^{-1}*R*P, corresponds to rotating back to the standard basis, doing your desired rotation R in the standard basis, then undoing the first rotation, with the net effect that you've done the desired rotation around j'. Since P is an orthogonal matrix, it follows that P^{-1} = P^T, so computing that matrix inverse is pretty easy.
Thanks for this. I have some things to think about.
So part of the issues I'm dealing with is my camera needs to be in uvn coordinates, which i pretty sure is the same as being in ijk. I've been looking around and found this matrix computation but I'm not quite getting it. I think if I understood this better I'd know what I need to do.
So I have a built in camera function(called glulookat) that I need to use. The camera works like this.
In the function I have 9 arguments. The first three dictate the location of C, or the eye. The second is the location of T or the center and the third is the location of the up vector. In this case that would be u. apparently uvn is left handed so I think that is correct. So, these finals values will all be in world coordinate values. I need to transform them from some uvn coordinate to said world coordinate. I've seen this calculation in a few places and want to go over it.
The first matrix is the basis matrix of the uvn coordinate. I'm not sure if this is in world coordinates or in its own, so it could be 1, 0, 0 for u but not be the worlds 1, 0 ,0? I'm also confused with what the xyz world vector is supposed to be. Are we taking the basis vector with some random world vertex and getting what the camera coordination systems equivalent vertex would be? Can someone clarify this?
boviscopophobic
Member
OK, well, there's a lot that could be covered and I don't think it would be helpful or appropriate to go into complete detail over a message board. Since you're taking a graphics course, it might be helpful to ask your instructor, as other students may also be dealing with the same issues. But my brief comments are as follows.
Presumably the projection plane is supposed to be located at a fixed distance from the camera C. Thus if you know C, you can get T from n and vice versa. Guessing from your description of the glulookat function, you need to keep track of C and the u, v, n vectors in the world basis. Keeping track of u, v, n just involves keeping them updated when the camera rotates, which was discussed in the previous messages. To keep track of C, you need to update it appropriately when the camera translates. e.g. if the player presses 'w' to go forward 2.7 units, you update C <- C + 2.7n. If they press 'a' to strafe left 2 units, you update C <- C - 2u, etc. I would assume that the glulookat function handles the details of perspective projection of world coordinates onto the projection plane.
Ignoring the fourth dimension (the 4th row/column) in your second equation for a moment, it shows a change of basis from world basis to camera basis. If C is at the world origin, then transforming an object's location from world coordinates to camera coordinates is a linear transformation. The matrix A of the transformation is obtained by writing down the components of the u, v, n vectors in world coordinates. Then left multiplication by A takes the world coordinates of the object to the camera coordinates of that object.
If C is not at the origin, then the coordinate transformation is not linear, but is affine (a linear transformation plus a translation). The fourth row/column of the matrices in your equation are added on as a device that allows 3D affine transformations to be expressed as 4D linear transformations. In this case, the change of basis matrix will take on a somewhat different form than what appears in your equation. However, that's beyond the scope of this post.
Presumably the projection plane is supposed to be located at a fixed distance from the camera C. Thus if you know C, you can get T from n and vice versa. Guessing from your description of the glulookat function, you need to keep track of C and the u, v, n vectors in the world basis. Keeping track of u, v, n just involves keeping them updated when the camera rotates, which was discussed in the previous messages. To keep track of C, you need to update it appropriately when the camera translates. e.g. if the player presses 'w' to go forward 2.7 units, you update C <- C + 2.7n. If they press 'a' to strafe left 2 units, you update C <- C - 2u, etc. I would assume that the glulookat function handles the details of perspective projection of world coordinates onto the projection plane.
Ignoring the fourth dimension (the 4th row/column) in your second equation for a moment, it shows a change of basis from world basis to camera basis. If C is at the world origin, then transforming an object's location from world coordinates to camera coordinates is a linear transformation. The matrix A of the transformation is obtained by writing down the components of the u, v, n vectors in world coordinates. Then left multiplication by A takes the world coordinates of the object to the camera coordinates of that object.
If C is not at the origin, then the coordinate transformation is not linear, but is affine (a linear transformation plus a translation). The fourth row/column of the matrices in your equation are added on as a device that allows 3D affine transformations to be expressed as 4D linear transformations. In this case, the change of basis matrix will take on a somewhat different form than what appears in your equation. However, that's beyond the scope of this post.
Thanks for your time. I'm trying to get into office hours tomorrow or friday to discuss it more. He didn't teach us any of this so I guess he expected everyone to already know how to do it from linear algebra(which I took over two years ago and haven't used since).
For maybe a more simple question, I'm having trouble trying to grasp the n v u equation above. n and u are simple to understand but v is less so. We need to already know what up is, but since I need to be rotating about every axis I figure the up vector would be changing all the time as well, and in the end I would think its just the v vector, but then if that's the case then I'd only be able to calculate one of the three vectors since the up vector changes.
For maybe a more simple question, I'm having trouble trying to grasp the n v u equation above. n and u are simple to understand but v is less so. We need to already know what up is, but since I need to be rotating about every axis I figure the up vector would be changing all the time as well, and in the end I would think its just the v vector, but then if that's the case then I'd only be able to calculate one of the three vectors since the up vector changes.
you need to divide both sides by P before you can take the natural logarithm of both sides because P is not being raised to the nt power
or you could take the natural logarithm of both sides, use the product rule of logarithms (or whatever it's called) for the right side to seperate ln[P(1+r/n)^nt] into lnP + ln[(1+r/n)^nt] and go from there. you will end up with the same answer
Thanks, that did the trick. Though now I'm stumped again.
52 = 100[1 - (.9)^t]
My brains being dumb or something. But if I try to work this to take log, it ends up trying to take the log of a negative number, which isn't possible :/ What am I doing wrong?
If you have done everything correctly up to that step, then you should have negatives on _both_ sides of the equation. Then you can simply divide both sides by -1 to make things positive before taking the log. (By the way, you are ahead of most of my students by even realizing that you can't take the log of a negative.)
Bah, I should have seen that lol. Thanks for the help. Yeah, it's pretty simple to see why a log of a negative just wouldn't work lol.
MisterLuffy
Member
I have another question...well it's more of if my answers are correct or if I'm in the right track. Also, if I need to do more work than what I have now.
Use mathematical induction to prove the statements are correct
for n ∈ Z+(set of positive integers).
n^3+5n divisible by 6; prove for integers n > 0
Base step:
n = 1 => (1)^3 + 5(1) = 6/6 = 1; True
n = 2 => (2)^3 + 5(2) = 18/6 = 3; True
Inductive step:
p(k): k^(3) + 5k = 6m (I did this because I'm copying one example my instructor did in class)
p(k+1): (k+1)^(3) + 5(k + 1) = k^(3) + 3k^(2) + 3k + 1 + 5k + 5 = (k^(3) + 5k) + 3k^(2) 3k+ 6
= 6m + 3k^(2) + 3k + 6 = 3(2m + k^(2) + k + 2) = 3*r (I'm substituting the term for r, again copying the example my instructor did in class)
3*r, r∈ Z+
=>Inductive step proven: n^(3) + 5n = 6m, n, m∈ Z+
Use mathematical induction to prove the statements are correct
for n ∈ Z+(set of positive integers).
n^3+5n divisible by 6; prove for integers n > 0
Base step:
n = 1 => (1)^3 + 5(1) = 6/6 = 1; True
n = 2 => (2)^3 + 5(2) = 18/6 = 3; True
Inductive step:
p(k): k^(3) + 5k = 6m (I did this because I'm copying one example my instructor did in class)
p(k+1): (k+1)^(3) + 5(k + 1) = k^(3) + 3k^(2) + 3k + 1 + 5k + 5 = (k^(3) + 5k) + 3k^(2) 3k+ 6
= 6m + 3k^(2) + 3k + 6 = 3(2m + k^(2) + k + 2) = 3*r (I'm substituting the term for r, again copying the example my instructor did in class)
3*r, r∈ Z+
=>Inductive step proven: n^(3) + 5n = 6m, n, m∈ Z+
fannybandit
Neo Member
it hasn't been proven. you only proved that p(k+1) is divisible by 3. you have to show that p(k+1) = 6*n where n is an integer.
What I was about to post. Also, since you've shown that
p(k+1) = 6m + 3k^2 + 3k + 6 = 6(m+1) + 3k^2 + 3k,
you've reduced the problem to showing that 3k^2 + 3k is divisible by 6, which is not difficult to do.
MisterLuffy
Member
Thanks for your help.....
Piercedveil
Member
Can anyone help me set up a differential equation for this:
"Water flows into a tank at the variable rate, Rin = 25/(1+t) and out at the constant rate Rout = 5 gal/min. Solve it with the initial condition V(0) = 110."
What I was thinking was the equation would be something like:
dV/dt = 25/(1+t) - 5V
But that doesn't seem quite right and I feel like I'm missing something (5V should be divided by something maybe?). I'm having a hard time figuring out how to set it up since the problem doesn't tell me how much the tank can hold like the other questions I've been doing.
Thanks in advance for any help!
"Water flows into a tank at the variable rate, Rin = 25/(1+t) and out at the constant rate Rout = 5 gal/min. Solve it with the initial condition V(0) = 110."
What I was thinking was the equation would be something like:
dV/dt = 25/(1+t) - 5V
But that doesn't seem quite right and I feel like I'm missing something (5V should be divided by something maybe?). I'm having a hard time figuring out how to set it up since the problem doesn't tell me how much the tank can hold like the other questions I've been doing.
Thanks in advance for any help!
Earthstrike
Member
What exactly does the question want you to solve?
Piercedveil
Member
All it wants me to do is "Set up the equation using the information given and solve using V(0) = 110."
So looks like it wants me to:
- set up the equation (do all the integration and such and get to V = (something * t) + C)
- solve for C (from the integral) using V(0) = 110.
and then answer with my result (an equation)
Sorry if it's a little confusingly worded. I'm pretty much restating exactly what it's saying, and it's confusing me too lol.
A question like this would normally ask you to write V as a function of t, which I assume is what's wanted here. dV/dt should just be the rate of flow in minus the rate of flow out. For this problem you can plug those values in exactly as given, i.e.
dV/dt = 25/(1+t) - 5.
The rate out doesn't depend on the current volume of liquid at all, so V doesn't need to show up there.
Piercedveil
Member
higgsboson1997
Member
Is math your favourite subject guys? It is mine!
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