2. There comes a time in everyone’s life when we have to solve for something, but there aren’t any equations readily available to use. Example: We’re told to find the time it takes for a runner to travel 100 meters while running at a speed of 10 meters per second. We have an equation that relates distance, speed, and time: often read as “distance = rate × time”. We can plug in the numbers we’re given into the equation: Unfortunately, what we want (time) is not by itself on one side of the equal sign; we have to isolate it, and then we’re able to solve. We can just divide both sides by 10 here, and we get that it takes the runner 10 second to run the distance. Instead, we have to manipulate equations we have in order to get what we want.

3. Sometimes, it’s easier to solve for the variable we want, and then plug in the numbers we’re given. In the previous example, we want to find the time. If we solve our equation first for time, we get: Since we’re looking for time, I’m dividing both sides by r to isolate the t. Now, we can plug in our distance and rate values we are given to find the time: just like before.

4. The previous example showed two ways to solve for what we’re looking for: either by plugging numbers in and then solving, or isolating our variable first and then plugging in our given values. Honestly, neither way was easier than the other. However, some equations are much more involved, and if we plug in numbers first we run the risk of making mistakes. Equations that almost require solving for different variables include: Square-root-type equations: Logarithmic functions: Trig-related functions: Exponentials: Of course these are extreme examples, but it’s important to see that we can manipulate even super-gross-looking equations to get what we want. Root-mean-square speed of particles in a gas, where k is a constant, T is temperature, and m is mass of a molecule Exponential growth/decay, where C, k are constants and t is time Distance modulus, where m, M are the apparent and absolute magnitudes of a star, and d is the distance

5. Let’s do a problem involving one of those hideous equations on the previous slide! What temperature is required for Neon to have a root-square-mean velocity of 1500 m/s? The previous slide says that k is a constant (it’s known as Boltzmann’s Constant), m is the mass of one molecule, and T is the temperature (which is what we’re solving for). Plugging in numbers is going to be nasty! We have our T term inside a square root, and just trying to get numbers first and then solving for the temp will set us up for disaster for sure. To solve for T, we just treat everything else as though it’s a number (I know everything except 3 is represented as a letter, but let’s give it a go anyway). What should we do first? It looks like we need to ditch the square root symbol. To get rid of that, we need to square both sides of our equation: 2

6. Now it’s time to solve for T : This is in Kelvin, so converting to degrees Fahrenheit gives us a temperature of -457.304 °F. That’s cold!

7. Solve the following equation for time: x = x 0 + v 0 t + ½ at 2 We want to solve for t, but unfortunately we can’t easily isolate t ; we have a regular t term as well as a t 2 term. What should we do? So, it looks like our best bet for this is to use the quadratic formula: Without actual numbers, we can’t do anything more to this equation. We’re done!

8. A certain isotope has a half-life of 500 years. If we start with 10g of the material, how much will be left after 329 years? This question involves some sort of decay. So, we need to use that equation for exponential growth/decay: Here, N is the amount of our material we have, C is our starting amount, and k is what’s known as the growth/decay constant. Where on Earth do we start?? It appears that we’re missing a value to solve this problem: We’re given a time and an amount, but we don’t have any clue what that growth constant should be. Actually, it’s not that much of an issue. All we have to do first is solve for that k value, and then we can answer the question this example wants us to do.

9. The secret to solving for k lies in the half-life part. When something has a half-life of 500 years, that means that after 500 years, we will be left with half as much of the stuff than when we started. So, if we start with 10g of the isotope, then we can expect to have 5g of it left after we let it sit around for five centuries. But first, let’s solve for k : Let’s move the C term on the other side of the equation: Now we’re stuck with some sort of strange e-to-the-something term! How are we going to get rid of that? That “e-to-the-something” is called an exponential function. In order to get rid of exponentials, we need to take the logarithm of both sides: This ln thing is called the natural log function. e is a special number, so its logarithm gets to be special, too. Logarithms are what “undo” exponentials; it’s really similar to square roots “undoing” things to the second power, only this time, the stuff that’s in the power is actually our variable, so it’s not entirely the same. ⇒

10. Logarithms just “cancel out” the exponential, like this: Now, solving for k is much more straightforward: So k = –0.001 4. Our growth/decay constant is negative. This is perfectly fine, since a negative k value tells us our amount of stuff is decreasing as time goes on—exactly what it should be doing! Now we can plug this value of k into our original equation: N = Ce –0.001 4 t ⇒

11. Using t = 329 and C = 10, we have N = Ce – 4 t = 10 e –0.001 4 (329) = 10 e –0.4606 = 10(0.6309) = 6.309 Thus, we’re left with 6.309g of the material after 329 years. Does this make sense as an answer? Time to solve our original question: We have 10g of the isotope (that’s our C value), and we want to know how much of it is left after 329 years (that’s our t value). Looks like it’s just a matter of plugging in numbers now, since what we’re solving for is already by itself on the other side of the equal sign!