[HomeWork]exponential growth

[Zerrotolerance]

Here is the question:

Suppose that a population of bacteria triples every hour and starts with 600 bacteria.
(a) Find an expression for the number n of bacteria after t hours.
n(t) =

(b) Estimate the rate of growth of the bacteria population after 0.5 hours. (Round your answer to the nearest hundred.)
n'(0.5) = bacteria/hour

We are currently doing implicit differentiation and all of a sudden this problem comes out of nowhere. I believe it is an exponential growth problem, but we haven't touched much on it in class besides bringing it up. I am not sure which way I should look at doing this? If it's exponential growth I would imagine the answer to part a) would look something like this:

600e^nt where 3 might possibly be substituted in for n.

I really have no idea what to do because we're not there yet, but I have 1 homework problem on this in my section on how velocity helped lead to finding derivitives? I am totally lost. I put what I thought the answer for part a was in and was marked wrong. I can't fix it now, but it's bothering me that I can't figure it out. I assume for b) you just plug .5 in for time? Any advice? Thanks!

 

[tkhunny]

Whoa! Take a breath. Don't panic.

Why do you need to identify the problem before you start? Just give it a go. Think it through and attack it.

Doubles every hour. \(\displaystyle 2^{t}\) Where t is time, in hours. Done.

Starts at 600 and doubles every hour. \(\displaystyle B(t) = 600\cdot 2^{t}\) Done.

Check. \(\displaystyle t = 0, B(0) = 600\cdot 2^{0} = 600 \cdot 1 = 600\)
Check. \(\displaystyle t = 1, B(1) = 600\cdot 2^{1} = 600 \cdot 2 = 1200\)

Looks good.

You tackle the derivative.

Note: I realize many text books INSIST that you use the base "e". Quite obviously, I disagree with this prescription. If you simply MUST use \(\displaystyle e^{kt}\) instead of \(\displaystyle 2^{t}\), go ahead and do that. It is NOT different.

 

[mmm4444bot]

600e^nt where 3 might possibly be substituted in for n.

Not quite.

I'm going to use the symbol k instead of n because your assignment already uses n as the name of the function.

n(t) = 600 * e^[ln(3) * t]

Here is one way to find the definition for function n.

We start with the basic form that you used for exponential growth (writing symbol k instead of n):

n(t) = n_0 * e^(kt)

n_0 is the initial amount, and k is the parameter which governs the rate of growth. We need to find the value of k.

Substitute in the known values after 1 hour:

t = 1

n_0 = 600

n(1) = 1800

That gives:

1800 = 600 * e^[(k)(1)]

Solve for k:

e^k = 1800/600 = 3

k = ln(3)

Now we can write the definiton for function n:

n(t) = 600 * e^[ln(3) * t]

And, yes, after you determine the expression for n`(t), you substitute t = 3 and evaluate n`(0.5) for part (b).

Cheers ~ Mark


MY EDIT: Fixed typographical error n`(3) for part (b).

 

[Zerrotolerance]

Actually I was wondering if you guys would be kind enough to explain some thing to me that I have been mixing up.

I am a little confused about the laws of finding the derivitive of a constant to a power. Here are some easy examples below I was wondering if you could go over.

ex.

1)5^x This would be 5^x right?

2)e^x This is the same as the example above?

3)e^2x This would be 2e^2x, I think the law is e^kt is ke^kt

4)e^-x This would be -e^-x

5)5^4 This would be 0

6)e^5^x This would be e^5^x(5^x)(1)

7)5^e^x This I am lost on because I know you can use the ln to somehow make this easier.

8)e^pi This would be 0 because both are constants

9)e^(pi3^x) Don't know

Also, could you go over an example where is would be better to use the laws of exponential function to find the derivitives. Something complicated like:

y = (5x^4-e^5x(ln(3)))^(2/3)/((3x+1)^(5/4))

I don't even know if it could be used on that. Sorry i just want to make sure I understand everything I need for tomorrow. I understand what should be the hard stuff like the chain, product, and quotient rule and when to use them, but struggle with some minor things. I know all the derivitives of trig/inverse trig function, but the constant ones mix me up somelikes. Thanks!

 

[mmm4444bot]

1) 5^x This would be 5^x right? Not quite; I'll explain.

2) e^x This is the same as the example above?

Yes. Both (1) and (2) are exponentials, a Real number (constant) raised to the power of x.

Here is the rule for the derivative of a constant with exponent x:

f(x) = C^x

f`(x) = C^x * ln(C)



3) e^2x This would be 2e^2x, I think the law is [the derivative of] e^kt is ke^kt That's a "shortcut" formula.

This is correct.

Yet, we need to change the Order of Operations, in your typing.

With e ^ 2 x we have e^2 before multiplying by x, yes?

e^(2x) is multiplying first, then using the exponent 2x on the natural base.

Same with e^(kt), too, please.

I see the rule as:

f(x) = C^(k*x)

f`(x) = k * C^(k*x) * ln(C)


4) e^-x This would be -e^-x

Yes. Better form when typing math text is to avoid typing adjacent operators like ^-

f(x) = e^(-x)

f`(x) = -e^(-x)


5) 5^4 This would be 0

Clearly.

Now that you know C^x * ln(C) is the derivative of C^x and the other rule above, fix your answer to (1), and give those others another go.

If you finished the exercise in your original post, I'd like to see your result.

 

[Zerrotolerance]

1) 5^x

The answer would be 5^x ln(5).

3) e^2x This would be 2e^2x, I think the law is [the derivative of] e^kt is ke^kt That's a "shortcut" formula.

This is correct.

Yet, we need to change the Order of Operations, in your typing.

With e ^ 2 x we have e^2 before multiplying by x, yes?

e^(2x) is multiplying first, then using the exponent 2x on the natural base.

Same with e^(kt), too, please.

I see the rule as:

f(x) = C^(k*x)

f`(x) = k * C^(k*x) * ln(C)

I understand what your saying and I will think of it like that from now on, but i am a little confused about how I had the correct answer and yet your answer also multiplied it by ln(c) which I know is also correct, by why is it ok to drop that when you simplify? I guess I am just a little rusty on logarithmic functions? When I see a function like this and don't realize what to do I usually use the chain rule in which case I get e^(2x)(2) or 2e^2x.


6) e^5^x This is (e^5^x ln e)(5x ln 5)(1) I think I could also use the chain rule without ln and get( e^5^x)(5^x)(1). Maybe?

7)5^e^x This is (5^e^x ln 5)(e^x ln e)(1) Chain rule-- (5^e^x)(e^x)(1)

8)e^pi is 0 e is a constant and pi is a constant (even though we're taught not to think of it as a number)

9)e^(pi3^x) Hmm...this one is difficult. I have no idea. (e^pi3^x ln e)(This is the part I don't know because multiplying pi3 you get 0) totally lost on this one.

The last part of the original problem I couldn't get.

y = 600e^(ln(3)t) I think you pull the constant out and do what we are doing now? 600(e^(ln(3)t) ln e)(ln(3)+1/(3t)(3)) Totally guessing on that. I used the product rule for ln(3)t but I don't know if that's correct. I got ln(3)t+1/(3t)(3) or ln(3)t+1/t. A little lost. Also why did you say to plug in 3 when it asked for the value after .5 hours. Did you just read it wrong or do I need to plug in 3.

Anyways, thanks for all the help. I would be lost in this class right now without you. Let me know if I made more mistakes. Hopefully I am starting to get the hang of it.

 

[mmm4444bot]

1) 5^x

The [derivative is] 5^x ln(5)

This is correct.

i am a little confused about how I had the correct answer [on the derivative of e^x] and yet your answer also multiplied it by ln(c)

Perhaps, you're not "seeing" the natural base e, when you look at ln(c).

I mean, in the exercise e^x, the constant c is the Real number e.

ln(e) = 1

why is it ok to drop that when you simplify?

In effect, we are not dropping anything; we are multiplying by 1, and we don't bother writing the factor ln(e) because it's trivial.

 

[mmm4444bot]

6) e^5^x

e^5^x is the constant e^5 raised to the power of x.

In other words, C = e^5

f(x) = C^x

f`(x) = C^x * ln(C)

Therefore, the derivative is (e^5)^x * ln(e^5)

We can simplify this to 5 e^5^x because ln(e^5) is 5.


7) 5^e^x This is (5^e^x ln 5)(e^x ln e)(1)

Yes. And we do not need to show the factors of ln(e) and 1.

5^(e^x) e^x ln(5)


8) e^pi is 0

The derivative is zero, yes.

Who taught you that Pi is not a number?!


9) e^(pi3^x)

Your typing means: e^(Pi * 3^x)

That's the same as:

f(x) = (e^Pi)^(3^x)

f`(x) = (e^Pi)^(3^x) * ln(e^Pi) * 3^x * ln(3)

(Chain rule, in addition to rules previously discussed.)

ln(e^Pi) = Pi

So, the derivative simplifies to e^(Pi * 3^x) * 3^x * ln(3) * Pi

 

[mmm4444bot]

n(t) = 600 e^(ln(3)t)

I used the product rule for ln(3)t

That's overkill.

The derivative of a constant times t is just the constant.

The derivative of ln(3) * t is just ln(3).


why did you say to plug in 3 when it asked for the value after .5 hours.

That was a typographical error; I fixed it.

n(t) = 600 e^[ln(3)*t]

n`(t) = 600 e^[ln(3)*t] * ln(3)

I get a rate of 1100 bacteria per hour, after 30 minutes.

 

[Zerrotolerance]

Wow, it's starting to make sense, thank you. I was a little confused about the whole ln e being dropped and ln 5 not before I realized the ln(e)=1. So, it's not dropped, just doesn't need to be written. I also just went over the last problem with my TA so I understant that too. He said nothing like that will be on this test. He said that's the section that we will have on the next exam.