Half Life and Exponential Growth Question

[ardentmed]

Hey guys,

I need help with a few more questions for this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

For the first one, I took v(t) as 0 since that is when it would be stopping or changing directions. Then I solved for t to get:

t=3 and t=0

As for acceleration, I took a(t)=0 and found that t=3/2 at that point.

Therefore, both v(t) and a(t) have the same signs, so "speeding up" will occur.

Then I graphed the question and got an upwards parabola for v(t) and a straight line for (at) which intersect around x~.5

As for the second question, I assumed an initial amount of 100mg (to make the computations easier to handle) and used the exponential growth formula y=Ce^(kt) for half life. At this point, I was able to plug in (0,58) into the formula and solve for k, giving me:

k= -.181576 (or just ln(0.58)/3

Thus, t ~ 3.817 days (I'm not sure about this though, nor do I know if the significant figs are right. Is it just 2 significant figures?)

As for 2b, I just assumed a 10mg amount out of 100mg, which is exactly 10%, and calculated t via substitution.

Therefore, t= 3ln(.1)/ln(.58)

So,

t=12.68 days.


Thanks in advance.

 

[MarkFL]

For the first problem, while you have correctly identified the roots of the velocity function, this is not what part a) is concerned with. when the particle is speeding up, its acceleration is positive and when it is slowing down, its acceleration is negative. So, you want to find the root(s) of the acceleration function and compute the sign of the acceleration function on the resulting intervals. What do you find?

For the second problem, we may state:

\(\displaystyle 0.58A_0=A_0e^{-3k}\)

\(\displaystyle 0.58=e^{-3k}\)

Hence:

\(\displaystyle e^{-k}=(0.58)^{\frac{1}{3}}\)

And so we have:

\(\displaystyle A(t)=A_0(0.58)^{\frac{t}{3}}\)

To compute the time it takes for a sample to decay to a portion of the original, which we can denote by $0<D\le1$, we may write

\(\displaystyle D=(0.58)^{\frac{t}{3}}\)

\(\displaystyle \ln(D)=\frac{t}{3}\ln(0.58)\)

\(\displaystyle t=\frac{3\ln(D)}{\ln(0.58)}\)

Now, we can use this formula to answer the two questions:

a) $D=0.5$:

\(\displaystyle t=\frac{3\ln(0.5)}{\ln(0.58)}\approx3.8174\)

b) $D=0.1$

\(\displaystyle t=\frac{3\ln(0.1)}{\ln(0.58)}\approx12.6811\)

Good job with these. :D

 

[ardentmed]

For the first problem, while you have correctly identified the roots of the velocity function, this is not what part a) is concerned with. when the particle is speeding up, its acceleration is positive and when it is slowing down, its acceleration is negative. So, you want to find the root(s) of the acceleration function and compute the sign of the acceleration function on the resulting intervals. What do you find?

For the second problem, we may state:

\(\displaystyle 0.58A_0=A_0e^{-3k}\)

\(\displaystyle 0.58=e^{-3k}\)

Hence:

\(\displaystyle e^{-k}=(0.58)^{\frac{1}{3}}\)

And so we have:

\(\displaystyle A(t)=A_0(0.58)^{\frac{t}{3}}\)

To compute the time it takes for a sample to decay to a portion of the original, which we can denote by $0<D\le1$, we may write

\(\displaystyle D=(0.58)^{\frac{t}{3}}\)

\(\displaystyle \ln(D)=\frac{t}{3}\ln(0.58)\)

\(\displaystyle t=\frac{3\ln(D)}{\ln(0.58)}\)

Now, we can use this formula to answer the two questions:

a) $D=0.5$:

\(\displaystyle t=\frac{3\ln(0.5)}{\ln(0.58)}\approx3.8174\)

b) $D=0.1$

\(\displaystyle t=\frac{3\ln(0.1)}{\ln(0.58)}\approx12.6811\)

Good job with these. :D

Wow, thanks a ton for going through all of that work to confirm my answer. This is why I love this community. Thanks again.

 

[ardentmed]

For the first problem, while you have correctly identified the roots of the velocity function, this is not what part a) is concerned with. when the particle is speeding up, its acceleration is positive and when it is slowing down, its acceleration is negative. So, you want to find the root(s) of the acceleration function and compute the sign of the acceleration function on the resulting intervals. What do you find?

For the second problem, we may state:

\(\displaystyle 0.58A_0=A_0e^{-3k}\)

\(\displaystyle 0.58=e^{-3k}\)

Hence:

\(\displaystyle e^{-k}=(0.58)^{\frac{1}{3}}\)

And so we have:

\(\displaystyle A(t)=A_0(0.58)^{\frac{t}{3}}\)

To compute the time it takes for a sample to decay to a portion of the original, which we can denote by $0<D\le1$, we may write

\(\displaystyle D=(0.58)^{\frac{t}{3}}\)

\(\displaystyle \ln(D)=\frac{t}{3}\ln(0.58)\)

\(\displaystyle t=\frac{3\ln(D)}{\ln(0.58)}\)

Now, we can use this formula to answer the two questions:

a) $D=0.5$:

\(\displaystyle t=\frac{3\ln(0.5)}{\ln(0.58)}\approx3.8174\)

b) $D=0.1$

\(\displaystyle t=\frac{3\ln(0.1)}{\ln(0.58)}\approx12.6811\)

Good job with these. :D

As for the acceleration for part a, I got decreasing when less then 3/2 and increasing for values over t=3/2. Where should I go from here?

Thanks again.

 

[MarkFL]

Okay, we are given the position function:

\(\displaystyle s=2t^3-9t^2\)

And so the velocity is:

\(\displaystyle v=\frac{ds}{dt}=6t^2-18t\)

And thus the acceleration is:

\(\displaystyle a=\frac{dv}{dt}=12t-18=6(2t-3)\)

And so we see that the root of the acceleration function is at:

\(\displaystyle t=\frac{3}{2}\)

And because it is an increasing linear function, we know it is negative to the left of the root and positive to the right. So, we are in agreement here. Given the fact that negative acceleration implies "slowing down" and positive acceleration implies "speeding up," how would you interpret your result?

edit: Actually, we need to be looking at speed $|v|$, not velocity $v$...so what do you find?

 

[ardentmed]

Okay, we are given the position function:

\(\displaystyle s=2t^3-9t^2\)

And so the velocity is:

\(\displaystyle v=\frac{ds}{dt}=6t^2-18t\)

And thus the acceleration is:

\(\displaystyle a=\frac{dv}{dt}=12t-18=6(2t-3)\)

And so we see that the root of the acceleration function is at:

\(\displaystyle t=\frac{3}{2}\)

And because it is an increasing linear function, we know it is negative to the left of the root and positive to the right. So, we are in agreement here. Given the fact that negative acceleration implies "slowing down" and positive acceleration implies "speeding up," how would you interpret your result?

edit: Actually, we need to be looking at speed $|v|$, not velocity $v$...so what do you find?

Speeding up occurs from: (3/2, infinty)
Slowing down occurs from: (0, 3/2)

Am I on the right track? Thanks again.

 

[MarkFL]

Well, we actually need to use:

\(\displaystyle |v(t)|=\begin{cases}6t^2-18t, & t<0 \\ 18t-6t^2, & 0\le t\le3 \\ 6t^2-18t, & 3<t \\ \end{cases}\)

What do you find?

 

[ardentmed]

Well, we actually need to use:

\(\displaystyle |v(t)|=\begin{cases}6t^2-18t, & t<0 \\ 18t-6t^2, & 0\le t\le3 \\ 6t^2-18t, & 3<t \\ \end{cases}\)

What do you find?

The particle is slowing down at (3, infinity)

and

Speeding up at (0, 3/2)

Is that close? Thanks again.

 

[MarkFL]

Suppose we let:

\(\displaystyle f(t)=|v(t)|\)

What is $f^{\prime}(t)$?

 

[ardentmed]

Suppose we let:

\(\displaystyle f(t)=|v(t)|\)

What is $f^{\prime}(t)$?

$f^{\prime}(t)$? = 6t(t + 3)Thus, t=3.

 

[MarkFL]

$f^{\prime}(t)$? = 6t(t + 3)Thus, t=3.

Recall the piecewise definition I gave above...

 

[ardentmed]

Given that:

\(\displaystyle |v(t)|=\begin{cases}6t^2-18t, & t<0 \\ 18t-6t^2, & 0\le t\le3 \\ 6t^2-18t, & 3<t \\ \end{cases}\)

f′(t)=12t-18 for t<0
f′(t)=18-12t for 0<t<3
f′(t)=12t-18 for t<3

The first one, when substituting in an arbitrary value with the limitation in mind gives a negative velocity. The second once gives a positive velocity, while the third one gives a negative velocity.

 

[MarkFL]

Yes, you differentiated correctly...now you want to find the signs of these pieces within their domains. :D

 

[ardentmed]

Yes, you differentiated correctly...now you want to find the signs of these pieces within their domains. :D

So it should be:
Speeding up from (0,3) and slowing down from (-$\infty$,0)u(3,$\infty$). But since the function starts at t=0, it is slowing down from (3,$\infty$)

Is that correct?

Thanks again.

 

[MarkFL]

We have:

\(\displaystyle f^{\prime}(t)=\begin{cases}12t-18, & t<0 \\[3pt] 18-12t, & 0\le t\le3 \\[3pt] 12t-18, & 3<t \\ \end{cases}\)

For all three pieces, we have the same root, which is \(\displaystyle t=\frac{3}{2}\). This root is only in the middle domain, so we need only look at the signs of the first and last pices in their respective domains. The first piece is negative and the last piece is positive, so we know the object is slowing down on:

\(\displaystyle (-\infty,0)\)

and speeding up on:

\(\displaystyle (3,\infty)\)

Now we look at the middle piece. We may observe that its derivative is a negative constant, thus its root is at a local maximum for $f$, and thus we may conclude that the object is speeding up on:

\(\displaystyle \left(0,\frac{3}{2}\right)\)

and slowing down on:

\(\displaystyle \left(\frac{3}{2},3\right)\)

 

[ardentmed]

We have:

\(\displaystyle f^{\prime}(t)=\begin{cases}12t-18, & t<0 \\[3pt] 18-12t, & 0\le t\le3 \\[3pt] 12t-18, & 3<t \\ \end{cases}\)

For all three pieces, we have the same root, which is \(\displaystyle t=\frac{3}{2}\). This root is only in the middle domain, so we need only look at the signs of the first and last pices in their respective domains. The first piece is negative and the last piece is positive, so we know the object is slowing down on:

\(\displaystyle (-\infty,0)\)

and speeding up on:

\(\displaystyle (3,\infty)\)

Now we look at the middle piece. We may observe that its derivative is a negative constant, thus its root is at a local maximum for $f$, and thus we may conclude that the object is speeding up on:

\(\displaystyle \left(0,\frac{3}{2}\right)\)

and slowing down on:

\(\displaystyle \left(\frac{3}{2},3\right)\)

Thank you so much for the thorough explanation. However, I still don't quite grasp why graphing the acceleration and velocity functions and finding which parts are both positive and which are negative doesn't work. What is the flaw in my logic?

Thanks again.

 

[Dethrone]

I haven't followed up on this question, but bear in mind, a particle is only speeding up when both its velocity and acceleration are in the same direction either both positive, or both negative. A particle is slowing down when the velocity and acceleration are in different direction, one is positive and the other is negative. I'm not sure what criteria you are using to determine whether it's slowing down or speeding up on the graph, but it will have to agree with my above statements.

- - - Updated - - -

You can see the four scenarios in the link below:
http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII_files/image001.gif

Note: Concave up = positive acceleration, concave down = negative acceleration, increasing = positive velocity, decreasing = negative velocity.

1. "Concave up, decreasing" - if you look at the picture I linked above, this is a situation where the velocity is negative and acceleration is positive. It is "slowing down" because the graph is getting less steep.

2. "Concave up, increasing" - velocity is positive and acceleration is positive, the graph is "speeding up" because it gets more steep.
3. "Concave down, decreasing" - velocity is negative and acceleration is negative, the graph is "speeding up" because it gets more steep.
4. "Concave down, increasing" - velocity is positive, acceleration is negative, the graph is slowing down because it gets more steep.

These are the criterion that you should be using when inspecting the graph to determine whether it is slowing down or speeding up.

 

[MarkFL]

I haven't followed up on this question, but bear in mind, a particle is only speeding up when both its velocity and acceleration are in the same direction either both positive, or both negative. A particle is slowing down when the velocity and acceleration are in different direction, one is positive and the other is negative. I'm not sure what criteria you are using to determine whether it's slowing down or speeding up on the graph, but it will have to agree with my above statements.

Yes, very good information, and such an analysis leads to the same results I gave, but in a more straightforward fashion. :D