Wed Oct 05, 2005 5:49 pm
CWisgood wrote:I've got a question too. It's from a maths test and our teacher and the whole class have different answers.
If ab<0, which of the following must be true?
i) a>0, b<0
ii) a<0, b>0
iii) a does not equal to 0, b does not equal to 0
iv) The signs (positive or negative) of a and b are diffirent.
Wed Oct 05, 2005 9:56 pm
shapu wrote:To "wax" is to emotionally say something. If I were talking about the woman of my dreams, I would "wax poetic." Use of the word "wax" requires that you use the adjectival form of the noun that follows it. So, rhapsody becomes rhapsodic, poetry becomes poetic, and so on.
The root of this usage comes from lunar cycles - when the moon is going from new to full, it is waxing, and when it is going from full to new, it is waning.
"Superiority of wild game to anything wrapped in cellophane" is the father ranting about how much he hates TV dinners, I believe. I haven't read the story you're referring to, but my guess is that he's trying to convince his daughter that hunted food is better than McDonald's.
As for "Sacred" - many ameroindian cultures viewed their relationship with their prey as a balance - the deer gave man meat, so man viewed the deer with respect.
Thu Oct 06, 2005 2:27 am
Matt wrote:CWisgood wrote:I've got a question too. It's from a maths test and our teacher and the whole class have different answers.
If ab<0, which of the following must be true?
i) a>0, b<0
ii) a<0, b>0
iii) a does not equal to 0, b does not equal to 0
iv) The signs (positive or negative) of a and b are diffirent.
To prove that something is wrong, the simplest way is to find a counter-example.
i) a=-1; b=1
=> ab=-1 .'. ab<0. However, a is not > than 0, so this is wrong.
ii) a=1; b=-1
=> ab=-1 .'. ab<0. However, a is not < than 0, so this is wrong.
iii) Let us assume that a=0. .'. ab=0 .'. ab is not < than 0. Let us assume that b=0. .'. ab=0 .'. ab is not < than 0. Therefore a and b cannot be 0.
iv) This is common sense, but I guess an axiom (Unprovable obvious thing.) I'll play about and see if I can get a proof.
You are right 3 and 4 are correct.
Thu Oct 06, 2005 3:52 am
M. Bison wrote:...
As you can see, lim h->0- =/= lim h->0+, and therefore, the limit does not exist, and consequently, the function is not differentiable at x = 0.
Let me know if this doesn't clear things up.
Sat Oct 08, 2005 1:36 pm
Sat Oct 08, 2005 8:57 pm
Matt wrote:Just a question:
Are you allowed to reciprocate both sides of an equation?
Ditto about inequalities.
Sun Oct 09, 2005 11:07 pm
A medicine is administered to a patient. The amount, M, of the medicine, in milligrams, in 1 mL of the patient's blood t hours after the injection is given by M(t) = -1/3t^2 + 1, where o < t < 3.
a) Find the rate of change of the amount, M, 2 h after the injection.
b) What is the significance of the fact that your answer is negative?
Mon Oct 10, 2005 1:05 am
Ammer wrote:A medicine is administered to a patient. The amount, M, of the medicine, in milligrams, in 1 mL of the patient's blood t hours after the injection is given by M(t) = -1/3t^2 + 1, where o < t < 3.
a) Find the rate of change of the amount, M, 2 h after the injection.
b) What is the significance of the fact that your answer is negative?
I need help, how do I do these questions?
Mon Oct 10, 2005 5:32 pm
M. Bison wrote:Ammer wrote:A medicine is administered to a patient. The amount, M, of the medicine, in milligrams, in 1 mL of the patient's blood t hours after the injection is given by M(t) = -1/3t^2 + 1, where o < t < 3.
a) Find the rate of change of the amount, M, 2 h after the injection.
b) What is the significance of the fact that your answer is negative?
I need help, how do I do these questions?
The (instantaneous) rate of change of a function is the derivative of the function.
Differentiate M(t) to find its derivative, denoted M'(t) and pronounced "M prime of t". You will probably just want to use the power rule for this.
M'(t) = -2/3t
It asks you for the rate of change (derivative) after 2 hours. Thus:
M'(2) = -4/3
A negative derivative at a point t indicates that the function is decreasing at t. Thus, the significance in this case is that it indicates the amount of medicine in the person's blood is decreasing at t = 2 hours (and, in fact, for all values of t such that 0 < t < 3).
Mon Oct 10, 2005 8:07 pm
Ammer wrote:M. Bison wrote:Ammer wrote:A medicine is administered to a patient. The amount, M, of the medicine, in milligrams, in 1 mL of the patient's blood t hours after the injection is given by M(t) = -1/3t^2 + 1, where o < t < 3.
a) Find the rate of change of the amount, M, 2 h after the injection.
b) What is the significance of the fact that your answer is negative?
I need help, how do I do these questions?
The (instantaneous) rate of change of a function is the derivative of the function.
Differentiate M(t) to find its derivative, denoted M'(t) and pronounced "M prime of t". You will probably just want to use the power rule for this.
M'(t) = -2/3t
It asks you for the rate of change (derivative) after 2 hours. Thus:
M'(2) = -4/3
A negative derivative at a point t indicates that the function is decreasing at t. Thus, the significance in this case is that it indicates the amount of medicine in the person's blood is decreasing at t = 2 hours (and, in fact, for all values of t such that 0 < t < 3).
First off, thanks for helping me.
But we haven't learnt derivative's yet and the answer in the textbook says -1/3. It is decreasing however the answer is different from what you got.
Mon Oct 10, 2005 8:41 pm
Matt wrote:Ammer wrote:M. Bison wrote:Ammer wrote:A medicine is administered to a patient. The amount, M, of the medicine, in milligrams, in 1 mL of the patient's blood t hours after the injection is given by M(t) = -1/3t^2 + 1, where o < t < 3.
a) Find the rate of change of the amount, M, 2 h after the injection.
b) What is the significance of the fact that your answer is negative?
I need help, how do I do these questions?
The (instantaneous) rate of change of a function is the derivative of the function.
Differentiate M(t) to find its derivative, denoted M'(t) and pronounced "M prime of t". You will probably just want to use the power rule for this.
M'(t) = -2/3t
It asks you for the rate of change (derivative) after 2 hours. Thus:
M'(2) = -4/3
A negative derivative at a point t indicates that the function is decreasing at t. Thus, the significance in this case is that it indicates the amount of medicine in the person's blood is decreasing at t = 2 hours (and, in fact, for all values of t such that 0 < t < 3).
First off, thanks for helping me.
But we haven't learnt derivative's yet and the answer in the textbook says -1/3. It is decreasing however the answer is different from what you got.
Is it possible that he misread is own working, and instead of (2/3)t, he actually wrote 2/(3t), in which case your answer would be correct?
Note that this is just a possibility, I actually don't now the calculus needed to work it through
Mon Oct 10, 2005 9:34 pm
M. Bison wrote:Matt wrote:Ammer wrote:M. Bison wrote:Ammer wrote:A medicine is administered to a patient. The amount, M, of the medicine, in milligrams, in 1 mL of the patient's blood t hours after the injection is given by M(t) = -1/3t^2 + 1, where o < t < 3.
a) Find the rate of change of the amount, M, 2 h after the injection.
b) What is the significance of the fact that your answer is negative?
I need help, how do I do these questions?
The (instantaneous) rate of change of a function is the derivative of the function.
Differentiate M(t) to find its derivative, denoted M'(t) and pronounced "M prime of t". You will probably just want to use the power rule for this.
M'(t) = -2/3t
It asks you for the rate of change (derivative) after 2 hours. Thus:
M'(2) = -4/3
A negative derivative at a point t indicates that the function is decreasing at t. Thus, the significance in this case is that it indicates the amount of medicine in the person's blood is decreasing at t = 2 hours (and, in fact, for all values of t such that 0 < t < 3).
First off, thanks for helping me.
But we haven't learnt derivative's yet and the answer in the textbook says -1/3. It is decreasing however the answer is different from what you got.
Is it possible that he misread is own working, and instead of (2/3)t, he actually wrote 2/(3t), in which case your answer would be correct?
Note that this is just a possibility, I actually don't now the calculus needed to work it through
Nope, -4/3 is the answer I intended to get. If you haven't learned derivatives yet, then you must have learned limits in order to properly answer this question. I am going to have to assume you have, because there really is no other way to answer this w/o limits and/or derivatives.
The derivative provides a mathematical formulation of rate of change; it measures the rate at which the function's value changes as the function's argument changes.
http://en.wikipedia.org/wiki/Derivative
Anyway, the derivative for a function, f(x) is defined as:
In this case, f(x) is really M(t), which equals -1/3t^2 + 1. I originally calculated the derivative with a shortcut known as the "power rule", but it can also be done the rigorous way using the limit-definition.
M'(t) = lim h->0 [M(t+h) - M(t)]/h
M(t) = -1/3t^2 + 1 and consequently, M(t+h) = -1/3*(t+h)^2 + 1.
M'(t) = lim h->0 [-1/3*(t+h)^2 + 1 - (-1/3t^2 + 1)]/h
"Foil" the (t+h)^2 to get t^2 + 2th + h^2:
M'(t) = lim h->0 [-1/3*(t^2 + 2th + h^2) + 1 - (-1/3t^2 + 1)]/h
Distribute everything:
M'(t) = lim h->0 [-1/3t^2 - 2/3*th - 1/3*h^2 + 1 + 1/3t^2 - 1]/h
Notice the -1/3t^2 and +1/3t^2 cancel, as does the -1 and + 1 (bolded above).
We are left with:
M'(t) = lim h->0 [-2/3*th - 1/3*h^2]/h
Factor out an h:
M'(t) = lim h->0 [h(-2/3t - 1/3h)]/h
Now notice the h on top and bottom cancel:
M'(t) = lim h->0 -2/3t - 1/3h
This is a limit, and as h approaches zero, -1/3h goes to zero, and we are left with:
M'(t) = -2/3t
Again, at t = 2 hours, M'(2) = -4/3. I don't mean to sound arrogant, but here's the truth: if your textbook gives you a different answer, it's wrong.
Wed Oct 26, 2005 3:52 pm
Wed Oct 26, 2005 9:31 pm
843 wrote:Someone please help me solve this Maths problem. I thought it was pretty simple at first but it actually is pretty tricky and I got stuck after solving to a certain extent...
Thu Oct 27, 2005 5:11 am