Professor Kate Poirier | OL67 | Fall 2020

Author: Brian Enriquez (Page 3 of 5)

How to Add Photos in Your Post

Seems like some people still have trouble uploading photos into their post so here’s a quick guide.

Click the plus sign (highlighted in yellow).
After clicking the plus sign, a pop up menu (highlighted in yellow) will appear. In the search bar of that pop up menu, type “image.”
Searching for image will show these options. Highlighted in yellow are options to add pictures. “IMAGE” only posts ONE picture. “GALLERY” posts MULTIPLE pictures in a “gallery.” Most of the pictures in this post used the “IMAGE” option.
After clicking either the IMAGE or GALLERY option, OpenLab will give you the option to upload your photos (highlighted in yellow).
Clicking the “upload button will show another pop up menu. Select the option where you can find your pictures.

Then your pictures should appear in your post.

OpenLab gives you the ability to write captions for your photos; just click on the image and you’ll see “write caption.”

Extra: below would be all the images from above compiled into a “gallery” if you chose the “GALLERY” option:

Project #1-Section 4.1-Exercise 1

Question:

The half-life of a radioactive substance is 3200 years. Find the quantity $Q(t)$ of the substance left at time $t > 0$ if $Q(0)=20$ $grams$?

Solution:

Context

The general exponential decay function is defined as:

$Q(t) = Q_0e^{-k(t-t_0)}$

$Q_0$  is the initial quantity, $-k$ is the “proportionality constant”, $t_0$ is the initial time, and $t$ would be any time duration.

For radioactive decay problems, $-k$ is treated as the “decay constant

Since $\tau$, or the “half-life,” is the amount of time at which a radioactive material’s quantity is reduced to half, we can turn $Q(t)$ into,

$Q(\tau) = \frac{Q_0}{2} = Q_0e^{-k(\tau-t_0)}$

$\frac{Q_0}{2} = Q_0e^{-k\tau}$

where $t_0 = 0$ and $t = \tau$.

We then solve for $k$ by canceling like terms and taking the natural logarithm of the equation:

$\ln(\frac{1}{2}) = \ln(e^{-k\tau})$

((Recall that $\ln(\frac{a}{b}) = \ln(a)-\ln(b)$ and $\ln(1) = 0$))

$-\ln(2) = -k\tau$

$k = \frac{ln(2)}{\tau}$

Actual solution

With these in mind, for Exercise 4.1.1, only algebra would be needed.

Given that $t_0 = 0$ and $Q(0) = 20$ $grams$,

$Q(t) = 20e^{-k(t-0)}$

And since $\tau = 3200$ $years$, solving for $k$ would yield

$k = \frac{ln(2)}{3200}$

Therefore, the quantity over time of a 20 gram substance with a 3200 years half-life can be found using,

$Q(t) = 20e^{-\frac{ln(2)}{3200}t}$

Note:
We don’t simply use $Q(t) = Q_0e^{-k\tau}$ or $Q(t) = 20e^{-k(3200)}$ as the solution because the resulting equation will NOT give us different values of $Q(t)$ at $t>0$, only at $t = \tau$. Remember that $\tau$ is just a value of $t$.

Solution by Brian and Jian Hui

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