Re: Distance Learning and Math Courses
@13, I actually have a scribe who will read me exams/quizzes and write down my work/answers. I practically pulled out all the stops when it came to figuring out accommodations - Braille book, scribe, notetaker, tutor. Fortunately, I have one single person who acts as all three which makes the process very smooth. The way you're doing it through Zoom is exactly how I do it ordinarily. I've been able to focus solely on the material without the worry of handling the accommodations in parallel. That being said, I worked on obtaining what I need up to a year in advance.
***The rest is really for the calc folks***
As far as implicit differentiation is concerned, it depends on with respect to what variable you are differentiating. Differentiating x^3 [with respect to x], it would be 3x^2 like you'd expect from the power rule. If you were differentiating y^3 [with respect to x], however, it would be 3y^2 * y-prime, where y-prime is dy/dx, or, the derivative of y with respect to x. All you're essentially doing is deriving all terms of an equation by multiplying by d/dx, so if you have:
x^3+y^3 = 8
and you were solving for dy/dx (y-prime)
Multiply all terms by d/dx (i.e.: take the derivative of each term with respect to x)
x^3 would become 3x^2 like above
y^3 would become 3y^2 * y-prime
8 would be 0 (from the rule of constants)
We're left with 3x^2 + 3y^2 * y-prime = 0
To solve for y-prime, subtract 3x^2 from both sides
Divide both sides by 3y^2 from both sides
y-prime = -3x^2/3y^2
The 3s cancel, so you're left with y-prime = -x^2/y^2
This will become far more apparent what this means and why it's important when covering related rates of change where you are differentiating with respect to t (time) and there is no explicit equation for solving for it, which is why we use implicit differentiation - we are implicating the rate at which A is changing with respect to B changing by determining the rate at which B is changing. For example, if a circle has a radius that is increasing at a rate of 3 cm per second, determine the rate at which the circumference of the circle is changing when the radius is equal to 5 cm. This is a pretty basic example but illustrates the concept as straightforward as I think you can get it. All you are being asked to find is the rate at which the circumference is increasing given two pieces of information: the length of the radius and the rate at which the radius is increasing. Since we have an equation for circumference, we know what r is, and we know the rate at which r is increasing, it's very simple to find the rate at which c (circumference) is increasing. Using c = 2PI * r, we [relating] the rate at which the radius is increasing, which is explicitly given to us by 3 cm per second to the other rate we are trying to find, in this case circumference. Note that the radius is increasing, so it is positive. It also turns out we don't need r = 10 for this example since deriving r just gives us 1. This makes sense because the circumference's rate of change is not strictly dependent on the radius itself. We just need to relate the radius's rate of change to the circumference's rate of change, so this example actually worked out better than I initially thought.
Remember, you're deriving every term in the equation you are using with respect to time. Since there is no variable for time (t), you will be implicitly differentiating.
The derivative of e^x is e^x per the chain rule, actually. I believe it's something like the derivative of a*e^x for some constant a is a * e^x * the derivative of the inside of the function, in this case, x. The derivative of x is just 1, so e^x*dx/dx = e^x.
PI * r^2 derived is 2PI * r with respect to r (the radius), so that checks out.
Some really cool stuff comes directly from mathematical concepts like exponential growth. Sparing the details that I'm guaranteed to mess up, first-order differential equations is actually the concept from which exponential growth is derived, but that's a Calc 2 concept.
What really made Calc 2 difficult is quite similar to @14 - there are a lot of different ways to integrate, which you get exposed to at the end of Calc 1 but really hammer in Calc 2. There's at least 6-7 ways to do so, and it's not always plainly apparent which method is the easiest/takes the least amount of steps to do. It is vitally important to keep up with all the different techniques since you'll be applying them left and right when learning other concepts. Limits also show up again and play a pretty big role at times, too.
***For everyone***
All of this to say is takes time to understand and practice. As @14 mentioned, make use of certain online resources like Khan Academy and OpenStax. Check out some YouTube videos. There's an English fellow who I'm blanking on the name that works. Not that these suggestions are exclusive to distance-learning circumstances, but are more emphasized due to that fact. I'd strongly suggest using them under normal circumstances as well.
Hope this helps!
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