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# Calculus Exam in 1 hour. New problem in OP

Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
Hi, H/A. As usual, I will be taking full advantage of our resident math gurus in my time of need. Thanks in advance!
First problem.
The population of Austria, P, in millions, is given by
P(t) = 5:6 (0.994)^t where t is time in years since 2000. How fast (in people/year) will it be changing in 2010?

I know that the answer is
P'(t) = 5.6 (0.994) ln (0.994)
= -31,733 people/year

I do not know where the ln (0.994) bit comes from. Is it some special property of the derivative of a number ^10? Something to do with the chain rule? Thanks again. I'll be adding problems as they come along.

Second problem.
The average cost per item to produce q items is given by a(q) = q^2 - 90q +3500. Find the minimum value of the marginal cost. (Hint: First find the total cost C(q), and then MC(q). Then minimize MC(q).)
So to get C(q) I would just multiply the a(q) function by q, since (average cost per unit) times (number of units) = (total cost), right?
That gives me C(q) = q^3 - 90q + 3500q.
Marginal cost would be the derivative, correct? C'(q) = 3q^2 - 180q + 3500
I believe the next step is to set the function equal to zero to get the critical point(s).
0 = 3q^2 - 180q + 3500
3500 = 3q^2 - 180q

And here I am stuck. Thanks again for the assistance.

TL DR on

## Posts

• Registered User regular
edited June 2009
Could 0.994 represent the overall change in population for one year, i.e. it decreases by 0.6%?

EDIT: Ahh, thought you meant the number 0.994 itself. I missed the ln() bit.

supertall on
• ___________PIGEON _________San Diego, CA Registered User regular
edited June 2009
supertall is wrong and it looks like you copied the answer wrong. I'm a HORRIBLE math teacher and a mediocre math student but this is simple so here goes:

The derivative of "a^x" is "(a^x)[ln(x)]." Most people just memorize this, but the reason is because of the chain rule: "a^x" is actually "e^[x ln(a)]." That is where ln(0.994) comes from.

TychoCelchuuu on
• Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
supertall wrote: »
Could 0.994 represent the overall change in population for one year, i.e. it decreases by 0.6%?

Yeah. 0.994 represents the change in population. However, if you just take P(10) = 5.6 (0.994)^10, you end up with about 5.273 . Compared to the original 5.6, that's a difference of -0.327 .

The answer I need to be arriving at is -0.031733

TL DR on
• Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
supertall is wrong and it looks like you copied the answer wrong. I'm a HORRIBLE math teacher and a mediocre math student but this is simple so here goes:

The derivative of "a^x" is "(a^x)[ln(x)]." Most people just memorize this, but the reason is because of the chain rule: "a^x" is actually "e^[x ln(a)]." That is where ln(0.994) comes from.

So the chain rule as I understand it is (derivative of outside function) (derivative of inside function) or f'(g(x)) times g'(x).

So I'd get something like
P(t) = 5.6(0.994)^t
P'(t) = 5.6 (0.994)^t ln(0.994)
Where ln(0.994) is the derivative of the ( )^10 function? Then substitute 10 and get
P'(10) = 5.6 (0.994)^10 ln(0.994)
= 5.24
???

TL DR on
• ___________PIGEON _________San Diego, CA Registered User regular
edited June 2009
No, it's not going to be ln(0.994), it's going to be ln(10), because d/dx of a^x is "(a^x)ln(x)", not "(a^x)ln(a)." I think. Honestly someone who has not gone years since taking calculus should be helping you, or at least someone who got an A when they took it years ago.

TychoCelchuuu on
• Get over yourself. Registered User regular
edited June 2009

Casually Hardcore on
• Registered User regular
edited June 2009
supertall is wrong and it looks like you copied the answer wrong. I'm a HORRIBLE math teacher and a mediocre math student but this is simple so here goes:

The derivative of "a^x" is "(a^x)[ln(x)]." Most people just memorize this, but the reason is because of the chain rule: "a^x" is actually "e^[x ln(a)]." That is where ln(0.994) comes from.
No, it's not going to be ln(0.994), it's going to be ln(10), because d/dx of a^x is "(a^x)ln(x)", not "(a^x)ln(a)." I think. Honestly someone who has not gone years since taking calculus should be helping you, or at least someone who got an A when they took it years ago.

Are you certain?

It seems to me that since a^x = e^[x ln(a)], it's far more likely that ln(a) will appear in the result than ln(x).

In fact, Wikipedia seems to think that the derivative is (a^x)ln(a), for a > 0.

Edit: Here we go:

(e^[ln(a) x]) ' = [ x ln(a) ] ' * e^[ln(a) x]
=> (a^x) ' = ln(a) * e^[ln(a) x] = ln(a) * (a^x)

ecco the dolphin on
• Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
I have this in my notes from when we did the review problem in class:
P'(t) = 5.6 (0.994)^t ln(0.994)
P'(10) = 5.6 (0.994)^10 ln(0.994)
=-0.031733 million people/year

It's possible I copied something down wrong, and I don't see how I got from P'(10) to my answer. We do have a copy of the practice test with answers though, and it supports the -0.031733 number.

TL DR on
• Registered User
edited June 2009
Remember "Pert" which is P0 x e^(rt) such that

P0 is the initial population at time zero (year 2000)
e is Euler's number (roughly 2.71, use e on your calculator for best results)
r is the rate (generally the unknown)
t is the time (10 years, referring to the period between 2000 and 2010)

Can you phrase the question this way? Otherwise I'm at a loss as to help you. I've only seen this type of question phrased like this.

Gaffero on
• Registered User regular
edited June 2009
Hi, H/A. As usual, I will be taking full advantage of our resident math gurus in my time of need. Thanks in advance!

First problem.
The population of Austria, P, in millions, is given by
P(t) = 5:6 (0.994)^t where t is time in years since 2000. How fast (in people/year) will it be changing in 2010?

I know that the answer is
P'(t) = 5.6 (0.994) ln (0.994)
= -31,733 people/year

I do not know where the ln (0.994) bit comes from. Is it some special property of the derivative of a number ^10? Something to do with the chain rule? Thanks again. I'll be adding problems as they come along.

Here's how I would do it (I know this is repeating and I think you know the right answer now, but I'll just write it out in hopes it solidifies it):

P(t) = 5.6 * (0.994)^t
= 5.6 * e^(t ln(0.994))

P'(t) = (5.6 * e^(t ln(0.994)))'

let f(t) = t * ln(0.994)
f'(t) = ln(0.994)

So P'(t) = (5.6 * e^f(t))'
= 5.6 * (e^f(t))' (since we can take constants out of derivatives)
= 5.6 * f'(t) * e^f(t) (by chain rule)
= 5.6 * ln(0.994) * e^(t ln(0.994))
= 5.6 * ln(0.994) * (0.994)^t

So P'(10) = 5.6 * ln(0.994) * (0.994)^10

Smug Duckling on
• Registered User regular
edited June 2009
So the chain rule as I understand it is (derivative of outside function) (derivative of inside function) or f'(g(x)) times g'(x).

So I'd get something like
P(t) = 5.6(0.994)^t
P'(t) = 5.6 (0.994)^t ln(0.994)
Where ln(0.994) is the derivative of the ( )^10 function? Then substitute 10 and get
P'(10) = 5.6 (0.994)^10 ln(0.994)

This is correct.

You must have typed something into your calculator wrong to end up with 5.24. If you evaluate that correctly, you get -0.0317328642 (thanks google calculator). That's in millions of ppl/year so you need to multiply by a million to get it into people/year.

Marty81 on
• Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
Ok, I think I have it now. Thanks so much for your patience. I'll be bumping this thread probably tomorrow before my exam if I need more help.

TL DR on
• Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
Ok, next problem.
The average cost per item to produce q items is given by a(q) = q^2 - 90q +3500. Find the minimum value of the marginal cost. (Hint: First find the total cost C(q), and then MC(q). Then minimize MC(q).)
So to get C(q) I would just multiply the a(q) function by q, since (average cost per unit) times (number of units) = (total cost), right?
That gives me C(q) = q^3 - 90q + 3500q.
Marginal cost would be the derivative, correct? C'(q) = 3q^2 - 180q + 3500
I believe the next step is to set the function equal to zero to get the critical point(s).
0 = 3q^2 - 180q + 3500
3500 = 3q^2 - 180q

And here I am stuck. Thanks again for the assistance.

TL DR on
• Registered User
edited June 2009
You need to differentiate again, then set C''(q) equal to 0. What you have is minimizing C, you need to minimize C'.

DJ-99 on
• Registered User regular
edited June 2009
that is right, marginal cost is C'. finding the minimum means another derivative

• Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
DJ-99 wrote: »
You need to differentiate again, then set C''(q) equal to 0. What you have is minimizing C, you need to minimize C'.

C''(q) = 6q - 180
q = 3

Then what, evaluate C'(3)?

TL DR on
• Registered User
edited June 2009
DJ-99 wrote: »
You need to differentiate again, then set C''(q) equal to 0. What you have is minimizing C, you need to minimize C'.

C''(q) = 6q - 180
q = 3

Then what, evaluate C'(3)?

Yes. Q=30 is the quantity, so to find the price plug back into your Marginal Cost function.

Hopefully you understand the general concept of minimizing functions (it seems like you mostly do but were a little confused here). If not, say so. But, to minimize C, you set C'=0. To minimize C', you set C''=0, etc.

EDIT: Q is actually 30. Think you just typed it wrong.

DJ-99 on
• Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
DJ-99 wrote: »
DJ-99 wrote: »
You need to differentiate again, then set C''(q) equal to 0. What you have is minimizing C, you need to minimize C'.

C''(q) = 6q - 180
q = 3

Then what, evaluate C'(3)?

Yes. Q=30 is the quantity, so to find the price plug back into your Marginal Cost function.

Hopefully you understand the general concept of minimizing functions (it seems like you mostly do but were a little confused here). If not, say so. But, to minimize C, you set C'=0. To minimize C', you set C''=0, etc.

EDIT: Q is actually 30. Think you just typed it wrong.

Yeah, twas just a typo. I got it. Thanks!

Next problem:
Minimize the cost of this fence. The area is 200 square feet, and the fencing along 3 sides costs $1/foot. The third side is$3/foot.

I have:
Area = 200
XY= 200
C = $1 (2x+y) +$3 (y)
C = 2x+4y
And we have to minimize Y.
So I guess something like
200/y = x
C = 2x + 4(800/x)
C = 2x + 3200/x
-3200 = 2x^2
-1600 = x^2
... I think I'm missing something. Do I want to make 3200/x into 3200 * x^-1 ?

TL DR on
• Registered User
edited June 2009
DJ-99 wrote: »
DJ-99 wrote: »
You need to differentiate again, then set C''(q) equal to 0. What you have is minimizing C, you need to minimize C'.

C''(q) = 6q - 180
q = 3

Then what, evaluate C'(3)?

Yes. Q=30 is the quantity, so to find the price plug back into your Marginal Cost function.

Hopefully you understand the general concept of minimizing functions (it seems like you mostly do but were a little confused here). If not, say so. But, to minimize C, you set C'=0. To minimize C', you set C''=0, etc.

EDIT: Q is actually 30. Think you just typed it wrong.

Yeah, twas just a typo. I got it. Thanks!

Next problem:
Minimize the cost of this fence. The area is 200 square feet, and the fencing along 3 sides costs $1/foot. The third side is$3/foot.

I have:
Area = 200
XY= 200
C = $1 (2x+y) +$3 (y)
C = 2x+4y
And we have to minimize Y.
So I guess something like
200/y = x
C = 2x + 4(800/x)
C = 2x + 3200/x
-3200 = 2x^2
-1600 = x^2
... I think I'm missing something. Do I want to make 3200/x into 3200 * x^-1 ?

You messed up. First of all, C = 2x + (800/x) NOT 3200/x.

Find C'. Set C'=0. Solve for x. Then plug X into xy=200 to solve for y.

DJ-99 on
• Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
I think I'm going to lose more points to careless mistakes than anything :|
Thanks for the help. Wish me luck.

edit: Oh, looks like the exam is actually at 4:30. Gonna review some more

TL DR on
• Registered User
edited June 2009
Good luck. If you have any spare time at the end just go back and double-check all your arithmetic. But, it is calculus, so understanding the concepts should at least gain you a bunch of points, even if you make a multiplication error.

DJ-99 on
• Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
f(x) = [(2x+1) / (3x-2)]^3
Find f'(1)

Can I evaluate this as f(1) and then derivate that, or should I find the derivative and then plug in 1 for x?

TL DR on
• Registered User
edited June 2009
f(x) = [(2x+1) / (3x-2)]^3
Find f'(1)

Can I evaluate this as f(1) and then derivate that, or should I find the derivative and then plug in 1 for x?

You have to find the derivative and then plug in. If you evaluated f(1) you would get a constant, and the derivative of that would be 0.

DJ-99 on
• Not at all confident in his reflexive opinions of thingsRegistered User regular
edited June 2009
DJ-99 wrote: »
f(x) = [(2x+1) / (3x-2)]^3
Find f'(1)

Can I evaluate this as f(1) and then derivate that, or should I find the derivative and then plug in 1 for x?

You have to find the derivative and then plug in. If you evaluated f(1) you would get a constant, and the derivative of that would be 0.

Hmm. I'm looking at something like
[2(3x-2) - 3(2x+1)] / (3x-2)^2
[(6x-4) - (6x+3] / (3x-2)^2 for the inside function via quotient rule. Correct? Then continue by taking (derivative inside function) (outside function) + (inside function) (derivative outside function) ?

TL DR on
• Registered User
edited June 2009
DJ-99 wrote: »
f(x) = [(2x+1) / (3x-2)]^3
Find f'(1)

Can I evaluate this as f(1) and then derivate that, or should I find the derivative and then plug in 1 for x?

You have to find the derivative and then plug in. If you evaluated f(1) you would get a constant, and the derivative of that would be 0.

Hmm. I'm looking at something like
[2(3x-2) - 3(2x+1)] / (3x-2)^2
[(6x-4) - (6x+3] / (3x-2)^2 for the inside function via quotient rule. Correct? Then continue by taking (derivative inside function) (outside function) + (inside function) (derivative outside function) ?

Not really sure what you're saying in the first line there. But I think what you want is this:

1. [(6x-4) - (6x+3] / (3x-2)^2 (this is what you have)
Multiply 1. by 3[(2x+1)/(3x-2)]^2 (you should see where this comes from).

I think this is right. At this point you can plug in a value of 1 for x and solve.

EDIT: It's been years since I did a differentiation this complex, so I could definitely be wrong.

DJ-99 on