Homework Help Thread

[Matt]

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.

[Divine]

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.

Thanks so much =). I understand now! (the wax thing was the thing that confused me the most, but I get it now =P)

[M. Bison]

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.

You might not even need (or be able) to prove IV, depending on the scope of the numbers we're talking about. If complex numbers are included:
a = b = i
ab = -1 < 0.

Strictly speaking, only III must be true.

[Yoshi]

...

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.

Sorry for the very late reply, but thanks a lot, it really helps. Derivatives I'll be learning next, and everything in your edit was pretty much in the method of my teacher. :)

[Matt]

Just a question:

Are you allowed to reciprocate both sides of an equation?

Ditto about inequalities.

[Divine]

Just a question:

Are you allowed to reciprocate both sides of an equation?

Ditto about inequalities.

I think you are. Basically you are just putting both sides of the equation to the power of -1.

You are allowed for inequalities too, but I believe you have to reverse the sign.

i.e.

1/2 < 1/x
2 > x (and x isn't equal to 0? I think...)

(I'm so sorry if this is wrong. I'm totally not sure of it)

Last edited by Divine on Mon Oct 10, 2005 1:14 am, edited 1 time in total.

[Ammer]

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?

[M. Bison]

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).

[Ammer]

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.

[Matt]

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

[M. Bison]

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.

[Ammer]

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.

Your way actually make sense.

I do believe my book made an error, and I will pass it on to my teacher.

Thanks a lot M. Bison! You really cleared it up!

[843]

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...

[M. Bison]

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...

You are given the length of OA, sqrt(250).

By the distance formula, sqrt(250) = sqrt((x2-x1)^2 + (y2-y1)^2). Of course, x1 = y1 = 0, so the formula is simply:

sqrt(250) = sqrt(x^2 + y^2) where x and y are the coordinates of the point A. Also, because we know y = 3x, we can eliminate y from the equation:

sqrt(250) = sqrt(x^2 + (3x)^2) = sqrt(10x^2)

Square both sides.

250 = 10x^2
25 = x^2
5 = x (To be painstakingly precise, in theory, -5 is also a possible answer, but we see from the graph that such an answer is clearly impossible).

We found x to be 5, and we know y = 3x, therefore the coordinate of point A is (5, 15).

Next, we know BA is perpendicular to OA, which has equation y = 3x. Thus, BA must have an equation of the form y = -1/3x + b (lines perpendicular to one another have negative inverse slopes). We don't know b (and if we did, we would know the coordinates of part B), but we sure can solve for it. We know the point (5, 15), or A, lies on line BA, thus substituting 5 for x and 15 for y yields:

y = -1/3x + b
15 = -5/3 + b
50/3 = b

Thus, point B must have coordinates (0, 50/3).

Finally, line CB is parallel to line OA, and must therefore have the same slope. Thus line CB can be expressed as y = 3x + b. Note that line CB has a y-intercept at point B, which we found to be 50/3. Thus CB's equation is y = 3x + 50/3.

To find point C, you simply need to find where CB and OC intersect (as you now know the equations of both). To do this, set them equal to each other:

3x + 50/3 = -2x
5x = -50/3
x = -10/3

Now that we have computed x, plug the value into one of the equations (let's take OC, y = -2x) to compute the y-coordinate.

y = -2x
y = -2*-10/3
y = 20/3

Thus, C has coordinates (-10/3, 20/3).

[843]

Wow, thanks a lot. That really helps

It actually is pretty simple, wonder why I was stuck...