Post by: Special At Specialist on February 08, 2012, 05:36:05 pm
ABCD is a quadrilateral in which AB is parallel to and 3 times the length of CD. If CD = a and AD = b then BC in terms of a and b is equal to:
A) 3b
B) b - 4a
C) -3b
D) 2a - b
E) 2a + b
I chose option B but apparently I'm wrong.
Post by: rife168 on February 08, 2012, 05:46:22 pm
A handy way to think about this is to trace the known vectors in order to get from B to C.
so (BA = 3a) + (AD = b) + (DC = -a)
therefore, 3a + b + (-a)
= 3a + b - a
= 2a + b
Post by: Special At Specialist on February 09, 2012, 03:33:04 pm
a, b and c are non-zero position vectors. The angle between vectors a and b is pi/4 and the angle between vectors a and c is pi/3. For the vectors to be linearly dependent, the angle between vectors b and c must be:
A) 5pi/12
B) 7pi/12
C) pi/2
D) pi/12 or 7pi/12
E) 5pi/12 or 7pi/12
I chose option A but I was wrong. Please help me solve this problem.
I think the answer was meant to be either D or E, but I'm not sure.
Post by: yawho on February 09, 2012, 09:06:59 pm
Post by: Special At Specialist on February 09, 2012, 09:34:26 pm
Post by: xZero on February 09, 2012, 10:17:39 pm
Post by: yawho on February 09, 2012, 10:50:41 pm
Post by: Special At Specialist on February 11, 2012, 12:32:39 pm
Now I have another problem:
Would we be expected to solve integrals like this in a tech free exam? How would I go about such a problem? I can't use substitution because there is nothing on the numerator and I can't use partial fractions since I cannot factorise the denominator.
Post by: pi on February 11, 2012, 12:42:51 pm
Post by: Special At Specialist on February 12, 2012, 04:19:27 pm
http://img830.imageshack.us/img830/3774/mathsproblem16.png
Moderator action: removed real name, sorry for the inconvenience
Post by: Special At Specialist on March 08, 2012, 01:46:54 am
Usually I would just do the dot product, but this is what happens:
a . b = -2 - 2 + 1
a . b = -3
|a| = sqrt(1 + 4 + 1)
|a| = sqrt(6)
|b| = sqrt(4 + 1 + 1)
|b| = sqrt(6)
a . b = |a||b|cos(θ)
-3 = 6cos(θ)
cos(θ) = -1/2
θ = 2pi/3 or θ = 4pi/3
But an acute angle is where 0<θ<pi/2. Please help me!
Post by: yawho on March 08, 2012, 10:00:03 am
Two vectors are given by a = -i - 2j + k and b = 2i + j + k. Find the acute angle between the vectors.The obtuse angle is 2pi/3, so the acute angle is pi - 2pi/3 = pi/3
Usually I would just do the dot product, but this is what happens:
a . b = -2 - 2 + 1
a . b = -3
|a| = sqrt(1 + 4 + 1)
|a| = sqrt(6)
|b| = sqrt(4 + 1 + 1)
|b| = sqrt(6)
a . b = |a||b|cos(θ)
-3 = 6cos(θ)
cos(θ) = -1/2
θ = 2pi/3 or θ = 4pi/3
But an acute angle is where 0<θ<pi/2. Please help me!
Post by: Special At Specialist on March 08, 2012, 05:41:05 pm
Post by: Lasercookie on March 08, 2012, 05:57:23 pm
How did you come to that conclusion? Did you move the vectors around?Just looking at the angles here. Maybe a diagram will help:
(http://i.imgur.com/vs2AN.gif)
If you look at the x-axis, 2pi/3 is the angle from the positive x-axis. So if you want to figure out the angle from the negative x-axis (the red bit in the diagram) and you know that the x-axis is a straight line (180 degrees, or pi radians), so pi - 2pi/3 would make sense.
I guess you can just think of it circular functions type stuff. 2pi/3 is in the second quadrant and then find the equivalent angle in the first quadrant.
Post by: Special At Specialist on March 28, 2012, 10:16:29 pm
f : [5, inf) -> R, f(x) = 2 + sqrt(x - 5)
So you can see from this function that the domain is [5, inf) and the range is [2, inf). But the range isn't written down; only the domain and co-domain are.
What is the co-domain? Why is it just R? Can you please give me an example of a function where the co-domain isn't R?
Post by: Lasercookie on March 28, 2012, 10:32:16 pm
What is a co-domain? I always see functions expressed something like this:The co-domain are the numbers that contain the range, while the range is the numbers that do come out of the function (in other words, the range is a subset of the co-domain). For the range [2, inf), the co-domain for that is Real (because the numbers between 2 and infinity are Real numbers).
f : [5, inf) -> R, f(x) = 2 + sqrt(x - 5)
So you can see from this function that the domain is [5, inf) and the range is [2, inf). But the range isn't written down; only the domain and co-domain are.
What is the co-domain? Why is it just R? Can you please give me an example of a function where the co-domain isn't R?
This function would have a co-domain of integers:
f: J --> J, where f(x) = 5x
Post by: Bhootnike on March 28, 2012, 10:38:08 pm
What is a co-domain? I always see functions expressed something like this:i think the co-domain is the set of possible values which may be produced, when the specific domain is applied to the equation. the range is the ACTUAL values which are produced. e.g.
f : [5, inf) -> R, f(x) = 2 + sqrt(x - 5)
So you can see from this function that the domain is [5, inf) and the range is [2, inf). But the range isn't written down; only the domain and co-domain are.
What is the co-domain? Why is it just R? Can you please give me an example of a function where the co-domain isn't R?
if dom was [1,4], and the codomain was [1,10], and your equation is, say, y=x+1,
then your range is, [2,3,4,5].
so codomain is like what may possibly be obtained, whereas range is the actual/obtained values you get when you 'plug' your x-values into the equation.
the codomain thus wont = R when you have:
f:[1,2,3,4,5] --> [1,2,3,4,5,6,7,8,9,10], f(x) = x+1
Post by: Special At Specialist on March 29, 2012, 04:14:31 pm
- The implied range
- The given range
- The codomain
And I have to find the values of the range which lie within all 3? And then to restrict it even further is the actual domain which will only produce certain values to begin with.
Post by: TrueTears on March 29, 2012, 05:11:40 pm
Post by: Bhootnike on March 29, 2012, 05:41:08 pm
So does that mean that the range is restricted by all 3 of these:
- The implied range
- The given range
- The codomain
And I have to find the values of the range which lie within all 3? And then to restrict it even further is the actual domain which will only produce certain values to begin with.
the range will definitely be restricted by codomain, and the actual domain as you've said, and i reckon your other points would be other restrictions too, (by 'given range' i guess you mean that they have specifically stated that the range is [a,b]) . implied domain - yeah makes sense, if you had,
just to be sure, someone more mathematical than me will have to confirm that !
Post by: Special At Specialist on April 03, 2012, 08:33:10 pm
You know how:
sin^2(x) = (sin(x))^2
sin^3(x) = (sin(x))^3
sin^-1(x) = arcsin(x)
Well what does sin^-2(x) equal?
Is it (sin(x))^-2 or is it (arcsin(x))^2 ?
Post by: brightsky on April 03, 2012, 08:46:14 pm
Post by: pi on April 03, 2012, 09:06:01 pm
Post by: Special At Specialist on May 24, 2012, 12:41:33 am
w = (4z + 9) / (z - 4) and z = w bar
If z = x + yi, show that (x - 4)^2 + y^2 = 25
Post by: Mao on May 24, 2012, 01:14:36 am
So, a circle centred at
Post by: Special At Specialist on May 24, 2012, 07:07:27 pm
If you have the equation x = 5, how do you find d(5) / dx ?
Would the answer be 1, since you treat 5 as the function of x and you take the derivative with respect to x, or would it be 0, since there is no change in the value of 5?
Post by: Mao on May 25, 2012, 02:23:58 am
Thanks Mao :) I have another question:The answer is undefined. X=5 is a vertical line. It's gradient would be +/- inifinity.
If you have the equation x = 5, how do you find d(5) / dx ?
Would the answer be 1, since you treat 5 as the function of x and you take the derivative with respect to x, or would it be 0, since there is no change in the value of 5?
Post by: Special At Specialist on May 25, 2012, 04:54:54 pm
Thanks Mao :) I have another question:The answer is undefined. X=5 is a vertical line. It's gradient would be +/- inifinity.
If you have the equation x = 5, how do you find d(5) / dx ?
Would the answer be 1, since you treat 5 as the function of x and you take the derivative with respect to x, or would it be 0, since there is no change in the value of 5?
Nooooo I'm not asking for the gradient. The gradient would be dy/dx.
I'm asking for d(5) / dx.
If you look at it from an algebraic perspective, you have 5 = x, therefore d(5) / dx = 1. But I'm not sure if it works that way which is what I'm asking.
But thanks for trying anyway.
Post by: kamil9876 on May 25, 2012, 05:59:17 pm
Post by: #1procrastinator on May 25, 2012, 07:02:15 pm
Post by: Special At Specialist on May 25, 2012, 07:22:11 pm
Hope I'm not hijacking the thread too much but what's the difference between differentiating functions and differentiating equations?
I wonder this too.
If I had an equation like:
e^y = 4x^2 - 8x and I want to find d(e^y) / dx, the answer would be:
d(e^y) / dx = 8x - 8
I understand that e^y is still a variable, not a constant, but would it be possible to differentiate with respect to a constant when the equation itself is a constant (eg. x = 5)?
Post by: Mao on May 26, 2012, 01:10:51 am
Thanks Mao :) I have another question:The answer is undefined. X=5 is a vertical line. It's gradient would be +/- inifinity.
If you have the equation x = 5, how do you find d(5) / dx ?
Would the answer be 1, since you treat 5 as the function of x and you take the derivative with respect to x, or would it be 0, since there is no change in the value of 5?
Nooooo I'm not asking for the gradient. The gradient would be dy/dx.
I'm asking for d(5) / dx.
If you look at it from an algebraic perspective, you have 5 = x, therefore d(5) / dx = 1. But I'm not sure if it works that way which is what I'm asking.
But thanks for trying anyway.
firstly, x=5 is not a function of x.
secondly,
Thirdly,
An alternative approach is if we were to take
Post by: Mao on May 26, 2012, 01:13:04 am
Hope I'm not hijacking the thread too much but what's the difference between differentiating functions and differentiating equations?
I wonder this too.
If I had an equation like:
e^y = 4x^2 - 8x and I want to find d(e^y) / dx, the answer would be:
d(e^y) / dx = 8x - 8
I understand that e^y is still a variable, not a constant, but would it be possible to differentiate with respect to a constant when the equation itself is a constant (eg. x = 5)?
e^y = 4x^2 - 8x <-- e^y is a function of x.
x = 5 <-- not a function of x. (if this can be written as a function of x, then this implies the constant 5 varies linearly with x, yet the constant 5 doesn't vary with anything, contradiction.)
Post by: Special At Specialist on May 27, 2012, 02:31:21 pm
I have another question:
A tank initially contains 200 litres of water. A sugar solution of concentration 30 grams per litre is added to the tank at a rate of 10 litres per minute. The mixture is kept uniform by stirring whilst being immediately drawn out at a rate of 7 litres per minute. If the tank contains x grams of sugar after t minutes, set up (but do not solve) a differential equation for x.
The problem with this question is that the rate of water flowing out is different to the rate of water flowing in. I've never attempted a problem like this before. Please help!
Post by: Special At Specialist on May 27, 2012, 03:36:39 pm
dV/dt = Input - Output
dV/dt = 10 - 7
dV/dt = 3
V = 3t + C
when t = 0, V = 200
200 = 3(0) + C
C = 200
V = 3t + 200
dx/dt = Input - Output
dx/dt = 10*30 - 7*(x / V)
dx/dt = 300 - 7x / V
dx/dt = 300 - 7x / (3t + 200)
Post by: Greatness on May 27, 2012, 04:53:36 pm
Post by: Special At Specialist on May 29, 2012, 10:15:42 pm
The graph of y = arccos(x), x E [-1, 1] is shown below.
(http://img694.imageshack.us/img694/5910/mathsproblem31.png)
a) Find the area bounded by the graph shown above, the x-axis and the line with equation x = -1.
b) Find the exact volume of the solid of revolution formed if the graph shown above is rotated around the y-axis.
For part a, I guessed that the area would be pi, but that is just a guess and I can't show any working out for it.
For part b, I know this involves comparing it to a y = cos(x) graph, I'm just not sure how, since they look quite different.
Post by: Greatness on May 29, 2012, 10:20:28 pm
b) isnt there a formula you learn for that? just plug the equation for y in? lol
Post by: TrueTears on May 29, 2012, 10:36:45 pm
Ugh this integration problem is really annoying me! It is question 9 from VCAA 2008 specialist mathematics exam 1:a) just intergrate along the y axis if you want to really show it.
The graph of y = arccos(x), x E [-1, 1] is shown below.
(http://img694.imageshack.us/img694/5910/mathsproblem31.png)
a) Find the area bounded by the graph shown above, the x-axis and the line with equation x = -1.
b) Find the exact volume of the solid of revolution formed if the graph shown above is rotated around the y-axis.
For part a, I guessed that the area would be pi, but that is just a guess and I can't show any working out for it.
For part b, I know this involves comparing it to a y = cos(x) graph, I'm just not sure how, since they look quite different.
Area from x = 0 to x = 1 is given by
Area from x = -1 to x = 0 is given by
Total area
Post by: Special At Specialist on May 31, 2012, 05:10:06 pm
If y = arccos(x) then x = cos(x)
Thus x^2 = cos^2(y)
since cos(2y) = 2cos^2(y) - 1
then cos^2(y) = (1/2)*(cos(2y) + 1)
V = pi*definite integral from y = a to y = b of x^2 dy
V = pi*definite integral from pi to pi/2 of (1/2)*(cos(2y) + 1) dy
V = pi/4 [sin(2y) + 2x] from pi to pi/2
V = pi/4 (0 + 2pi + 1 - pi)
V = pi/4 (pi + 1)
V = (pi^2 + pi) / 4
Post by: Lasercookie on May 31, 2012, 05:46:25 pm
Thanks for part a. I think I might have part b. Please tell me if this is right or wrong.All the above looks correct, you had the right idea for the integral too. (also latexified it, I find that a bit easier to read :P)
Ifthen
Thus
since
then
So we have x E [-1,1], so our bounds are y E [0, pi] (this can be just seen from the graph given, you made a small error here), written out anyway:
Post by: Special At Specialist on August 10, 2012, 05:24:26 pm
I thought the answer was (-∞, ∞), but other people keep saying it's [-pi/6, pi/6].
Could someone please explain?
Post by: Lasercookie on August 10, 2012, 06:42:20 pm
What is the implied domain of the function y = arccos(sin(3x))?I think you're right.
I thought the answer was (-∞, ∞), but other people keep saying it's [-pi/6, pi/6].
Could someone please explain?
The domain of arccos(x) is [-1,1]
The range of sin(3x) is [-1,1]
The domain of sin(3x) is R.
So you could input any real number into the function, and still get a number out.
Post by: Special At Specialist on August 10, 2012, 07:40:33 pm
Post by: Lasercookie on August 10, 2012, 07:43:59 pm
Hmm... does an arccos graph have to be a one-one function? Because if it does, then I understand how people got x E [-pi/6, pi/6] as the answer. If not, then I think it is x E (-∞, ∞).The domain of arccos is restricted to [-1,1] because it's a one-to-one function. What we have here is a composite function though, not just arccos by itself.
Post by: Hancock on August 10, 2012, 07:49:16 pm
What is the implied domain of the function y = arccos(sin(3x))?
I thought the answer was (-∞, ∞), but other people keep saying it's [-pi/6, pi/6].
Could someone please explain?
arccos' domain is restricted from [-1,1]. If that was an x in there instead of sin(3x) we would say that. However, we must now solve the inequality of -1 =< sin(3x) =< 1, because the x from arccos(x) has been replaced with sin(3x).
So:
-1 =< sin(3x) =< 1
arcsin(-1) =< 3x =< arcsin(1) (arcsin function output limited from -pi/2 to pi/2)
-pi/2 =< 3x =< pi/2
-pi/6 =< x =< pi/6
So the domain of arccos(sin(3x)) is [-pi/6, pi/6].
Post by: Special At Specialist on August 10, 2012, 07:56:32 pm
What is the implied domain of the function y = arccos(sin(3x))?
I thought the answer was (-∞, ∞), but other people keep saying it's [-pi/6, pi/6].
Could someone please explain?
arccos' domain is restricted from [-1,1]. If that was an x in there instead of sin(3x) we would say that. However, we must now solve the inequality of -1 =< sin(3x) =< 1, because the x from arccos(x) has been replaced with sin(3x).
So:
-1 =< sin(3x) =< 1
I got up to this part, and then I deduced that -∞ < x < ∞. Is there anything wrong with that logic?
Also, is this correct:
y = arccos(sin(3x))
since sin(u) = cos(pi/2 - u) then sin(3x) = cos(pi/2 - 3x)
y = arccos(cos(pi/2 - 3x))
y = pi/2 - 3x
Post by: Hancock on August 10, 2012, 08:01:07 pm
And as for the function F = pi/2 - 3x, it is equal to arccos(sin(3x)) only in it's defined domain of -pi/6 to pi/6. Good job on that one btw
Post by: Lasercookie on August 10, 2012, 08:19:28 pm
I can see for something like
However with this example of
Let's say we have
So
Post by: Hancock on August 10, 2012, 08:24:49 pm
I'm sorry to be stubborn, but I still don't have my head completely around this.
I can see for something likewe have to restrict the domain since the inequality
will not be true for all x. If you tried to substitute in
, then you'd end up with
which we know is outside the restricted domain of arccos(x).
However with this example of, and the inequality
we can let x be any real number, and we'll have a number that satisfies the inequality.
Let's say we have
I see what you mean, however the I believe that the inequality of
Post by: brightsky on August 11, 2012, 01:38:46 pm
@Hancock - how would you solve sin(3x) = 1? 3x = arcsin(3x) is only one solution. the inequality -1 =< sin(...) =< 1 always holds provided sin(...) has a large enough domain, which in this case it has.
Post by: Jenny_2108 on August 11, 2012, 02:22:19 pm
i'm pretty sure the domain is R. you can sub in any x and the function f(x)= arccos(sin(3x)) will spit out a value.
@Hancock - how would you solve sin(3x) = 1? 3x = arcsin(3x) is only one solution. the inequality -1 =< sin(...) =< 1 always holds provided sin(...) has a large enough domain, which in this case it has.
I think if we solve sin(3x)=1 => 3x+2npi=arcsin(1) (n is integers)
3x= pi/2-2npi
x=pi/6-2npi/3
Thus the implied domain is R.
If we put in CAS the function arccos(sin3x), the function is defined with all values of x
-1 =< sin(3x) =< 1
arcsin(-1) =< 3x =< arcsin(1) (arcsin function output limited from -pi/2 to pi/2)
-pi/2 =< 3x =< pi/2
-pi/6 =< x =< pi/6
So the domain of arccos(sin(3x)) is [-pi/6, pi/6].
If you say the range of arcsin function is -pi/2 to pi/2 so
arcsin(-1)=< 3x+2npi=<arcsin(1)
-pi/2-2npi=< 3x=<pi/2-2npi
-pi/6-2npi/3=<x=<pi/6-2npi/3
Therefore, the implied domain is R
Post by: Hancock on August 11, 2012, 03:04:08 pm
Post by: Special At Specialist on August 14, 2012, 05:02:18 pm
What is the domain of y = arcsin(sin(x)) ?
If I assume that y = x, then I end up with a straight line, rather than a zig-zag line. The line y = x is the same as y = arcsin(sin(x)) only in the domain [-pi/2, pi/2]. Everything outside that domain zig-zags up and down, whereas the y = x just continues in a straight line.
Post by: Lasercookie on August 14, 2012, 05:21:23 pm
From what I see, it seems all you're doing there is solving for
The implied domain is "the set of all real numbers for which the expression is defined.", and we can see that the function is defined outside of [-pi/2,pi/2]
If you're solving the inequality
With the not ignoring the other periods thing, it's also comes up with what you did with
Edit: You can look at an alternate problem, of a different form where the maximal domain would not be R, e.g.
Post by: Jenny_2108 on August 14, 2012, 07:55:41 pm
Considering the trig function, the maximal domain of arcsin(sinx) is still R because its defined with all values of x
But arcsin(sinx) is only equal x when x E [-pi/2,pi/2]
Outside this domain, arcsin(sinx) is still defined but not equal to x.
Does it make sense, Special at Specialist?
Post by: Special At Specialist on August 27, 2012, 05:52:20 am
Are these steps mathematically valid (ie. would they score full marks if written on a VCAA exam):
dy/dx = 2x
∫dy = ∫2x dx
y = x^2 + C
As opposed to:
dy/dx = 2x
y = ∫2x dx
y = x^2 + C
Basically, is it correct to split it up into 2 different integrals, or do I have to go from dy/dx = ... to y = ... dx?
Post by: #1procrastinator on August 28, 2012, 04:22:58 pm
But a question like that would be one mark, so do you really have to show the steps?
Post by: Special At Specialist on August 28, 2012, 04:34:57 pm
Post by: BubbleWrapMan on August 28, 2012, 06:39:10 pm
If it's a definite integral I do this:
Post by: Special At Specialist on August 28, 2012, 10:20:20 pm
Post by: pi on August 28, 2012, 10:24:22 pm
So then I was correct in saying ∫dy = ∫2x dx ? I wouldn't lose any marks for that, would I?
Nope
Post by: Jenny_2108 on August 28, 2012, 10:28:00 pm
So then I was correct in saying ∫dy = ∫2x dx ? I wouldn't lose any marks for that, would I?
Nope
Do we have to write step by step like this in exam? :O
Its just tedious and time-consuming
Post by: BubbleWrapMan on August 29, 2012, 11:17:50 am
Post by: #1procrastinator on August 29, 2012, 03:57:21 pm
Post by: Special At Specialist on August 29, 2012, 08:15:30 pm
Post by: Jenny_2108 on August 29, 2012, 08:17:38 pm
I don't think the step is necessary, but if I were to add it, then I wouldn't lose any marks for it.
why do you have to add it if its not neccessary?
Btw, I dont see people solve by writing as ∫dy = ∫2x dx though
Post by: pi on August 29, 2012, 08:19:00 pm
His step mentioned ∫dy = ∫2x dx is a perfectly valid one and I've seen it written by few in my school.
Post by: #1procrastinator on August 29, 2012, 08:32:35 pm
mean you're then integrating dy? Shouldn't you be treating dy/dx as a function and not be treating it as a fraction?
Post by: pi on August 29, 2012, 08:40:43 pm
Doesn't that
mean you're then integrating dy? Shouldn't you be treating dy/dx as a function and not be treating it as a fraction?
dy/dx obeys fraction laws. Perfectly legit.
Post by: #1procrastinator on August 29, 2012, 08:46:02 pm
Post by: pi on August 29, 2012, 08:56:43 pm
hmmm...just seems a bit like abuse of notation to me lol
Technically it kind of is if you are a stickler to proper notation and nothing less than that, but it's pretty well accepted by VCAA and by most mathematicians in general.
Post by: Hutchoo on August 29, 2012, 09:05:22 pm
hmmm...just seems a bit like abuse of notation to me lol
dy/dx is treated as a normal fraction. That's how the chain rule works.
Note: d^2y/dx^2 (or any type of differential that's higher than the first derivative cannot be treated as a fraction).
Post by: BubbleWrapMan on August 29, 2012, 10:11:30 pm
That's where the ∫dy comes from, so it's a valid step to write ∫dy = ∫f(x)dx after dy/dx = f(x), it just skips the derivation of it in the first line (above). So I guess if you write the second line you might as well write the first line. But yes, ∫dy is valid.
Post by: rife168 on August 30, 2012, 05:47:33 pm
hmmm...just seems a bit like abuse of notation to me lolYou're right:
http://en.wikipedia.org/wiki/Abuse_of_notation#Derivative
Post by: #1procrastinator on August 30, 2012, 06:29:20 pm
I wrote on the last page:
That's where the ∫dy comes from, so it's a valid step to write ∫dy = ∫f(x)dx after dy/dx = f(x), it just skips the derivation of it in the first line (above). So I guess if you write the second line you might as well write the first line. But yes, ∫dy is valid.
Yeah but I just can't get my head around what the the integral of nothing with respect to dy means cause it's like you killed the derivative function when you cancelled the dxs...but as I said, I'm not comfortable with differentials and if the examiners accept...well then that's fine :p
Post by: WhoTookMyUsername on August 30, 2012, 06:39:26 pm
dy/dx = f(u) x g(x)
is that "notationally correct"?
Post by: nerfsdacier on August 30, 2012, 06:43:56 pm
I wrote on the last page:
That's where the ∫dy comes from, so it's a valid step to write ∫dy = ∫f(x)dx after dy/dx = f(x), it just skips the derivation of it in the first line (above). So I guess if you write the second line you might as well write the first line. But yes, ∫dy is valid.
Yeah but I just can't get my head around what the the integral of nothing with respect to dy means cause it's like you killed the derivative function when you cancelled the dxs...but as I said, I'm not comfortable with differentials and if the examiners accept...well then that's fine :p
I think of it as integrating the number 1 with respect to y, which is why you get y as the result.
Post by: #1procrastinator on August 30, 2012, 07:00:32 pm
Post by: BubbleWrapMan on August 30, 2012, 10:35:49 pm
if you writeYes, if you define everything else properly when doing the question (I assume you had y = F(G(x)) and u = G(x) )
dy/dx = f(u) x g(x)
is that "notationally correct"?
Post by: WhoTookMyUsername on August 31, 2012, 05:28:45 pm
(as an intermediate step before subbing back in u=)
Post by: BubbleWrapMan on August 31, 2012, 10:27:06 pm
Post by: Jenny_2108 on August 31, 2012, 10:45:28 pm
if you write
dy/dx = f(u) x g(x)
is that "notationally correct"?
Do you mean:
And your question is: can you write as
Post by: WhoTookMyUsername on September 01, 2012, 09:12:34 am
an example is
let y = xloge(cosx))
and then you go let u = cos x then du/dx = -sinx
then y =xloge(u)
then
dy/dx = loge(u) x x(-sinx)/cos(x)
cos you have dy/dx in terms of both u and x
Post by: pi on September 01, 2012, 01:44:11 pm
Post by: paulsterio on September 01, 2012, 02:05:11 pm
nah
an example is
let y = xloge(cosx))
and then you go let u = cos x then du/dx = -sinx
then y =xloge(u)
then
dy/dx = loge(u) x x(-sinx)/cos(x)
cos you have dy/dx in terms of both u and x
pi is right - this is dodgy. Look at what you've done - from this step:
y =xloge(u)
to this step
dy/dx = loge(u) x x(-sinx)/cos(x)
you didn't use the product rule, what you should have done is:
dy/dx = d/dx(x) * loge(u) + x * d/dx(loge(u))
then you would have to use the chain rule in order to find the derivative of wrt to x of loge(u)
which would give you:
d/dx(loge(u)) = d/du(loge(u)) * d/dx(u)
...etc.
All in all - what you did is dodgy and not correct.
Moderator action: removed real name, sorry for the inconvenience
Post by: WhoTookMyUsername on September 01, 2012, 02:59:04 pm
It's whether daying dy/dx = f (u ) + g(x) is notationally correct
Like is saying dy/dx = ln (u) - xsin(x)/cos(x)
A valid ( notationally) intermediate step to
The final solution?
Dy/dx = ln(cox(x))-xtanx?
Like for integrals you can't go S 2x du
For example
But when you're deriving was my intermediate step (notationally) correct or incorrect?
Thanks
Post by: pi on September 01, 2012, 03:23:16 pm
Post by: Jenny_2108 on September 01, 2012, 03:43:14 pm
I don't think that's valid, just write the other intermediate step imo
Is the intermediate step compulsory?
I always do the shortcut but if I still get the right answer, do I lose marks for skipping steps?
Post by: pi on September 01, 2012, 03:53:42 pm
Post by: paulsterio on September 01, 2012, 04:07:38 pm
Moderator action: removed real name, sorry for the inconvenience
Post by: Jenny_2108 on September 01, 2012, 04:30:10 pm
I don't understand why these questions are asked
Whats wrong with asking questions?
Post by: pi on September 01, 2012, 04:33:13 pm
Post by: Jenny_2108 on September 01, 2012, 04:37:12 pm
It's questions where the answers should be fairly obvious tbh :P The general rule is to solving a question is to use the most elegant method and show the steps that will allow an examiner to easily and clearly see your flow of logic. The goal isn't to cut corners by making obvious dodgy steps (ie. those that aren't in textbooks or in practice exams).
I did the shortcut in SACs and it was fine, the teacher didnt take any marks off though
Thats why I ask whether I can do like this in exams
Anyway, I apologise for my silly question
Post by: paulsterio on September 01, 2012, 07:01:43 pm
Whats wrong with asking questions?
There's nothing wrong with asking questions, but, like pi, I think the answer is fairly obvious, you want to present your answers as best you can and there are plenty of resources out there including exams and textbooks which will show you how to present an answer.
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Post by: BubbleWrapMan on September 01, 2012, 07:22:55 pm
Post by: paulsterio on September 01, 2012, 07:30:57 pm
It's important for someone else to be able to follow your working, even outside VCE. Even if you don't go on with maths it's pretty much inevitable that you'll have to do some quantitative problem-solving later on, so it's good to practice being clear.
Good advice, and he's right, it's not just to do with Maths but in everything you do, from writing CVs to writing essays and delivering presentations, clarity is key!
Post by: Jenny_2108 on September 01, 2012, 07:55:33 pm
It's important for someone else to be able to follow your working, even outside VCE. Even if you don't go on with maths it's pretty much inevitable that you'll have to do some quantitative problem-solving later on, so it's good to practice being clear.
I was hoping that if I spend less time for easy questions or skip long steps, I can have more time for difficult ones
But yeah, you are correct. Not about Maths but other things as well, I should do carefully
Post by: BubbleWrapMan on September 01, 2012, 07:59:59 pm
Post by: WhoTookMyUsername on September 02, 2012, 08:58:14 am
is this the intermediate step?
wb for methods, what should you do there (can you use pseudo implicit diff?)
Post by: Special At Specialist on September 04, 2012, 08:20:58 pm
For example, one question asked me to resolve a vector into two components: one which was parallel to another vector and one which was perpendicular to it. I know that I have to use the vector resolute to find the parallel component and then subtract that from the first vector to find the perpendicular component, but I'm not sure how to show the working out, especially the final answer. Do I just write "parallel component = ... and perpendicular component = ..." ? Or is there another way of writing the answer?
Post by: paulsterio on September 04, 2012, 08:34:22 pm
It's not a matter of being too lazy to write the working out, it's knowing what to write. Often I know exactly how to solve a problem, I just don't know how to show the working out and I end up losing marks on such an easy question that it frustrates me.
For example, one question asked me to resolve a vector into two components: one which was parallel to another vector and one which was perpendicular to it. I know that I have to use the vector resolute to find the parallel component and then subtract that from the first vector to find the perpendicular component, but I'm not sure how to show the working out, especially the final answer. Do I just write "parallel component = ... and perpendicular component = ..." ? Or is there another way of writing the answer?
My advice is to be clear and logical, as long as you are clear and logical, you should be OK.