20181222, 18:13  #133 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
One thing that was on the test that was weird was derivatives of powers of trig functions. Here’s some work I’ve done, I’m not quite sure how you’re supposed to simplify it normally.

20181223, 09:34  #134  
Dec 2012
The Netherlands
5×349 Posts 
Quote:
\[ \frac{d}{dx}(\sin x)^3=3(\sin x)^2\frac{d}{dx}\sin x=3(\sin x)^2\cos x. \] 

20181223, 22:47  #135 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}×3×37 Posts 
Oh, yeah. Duh!

20181224, 16:36  #136 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}×3×37 Posts 
Here’s another one that was weird. I’ve probably gotten the wording wrong, since it was worded poorly in the first place and I can’t remember things anyways
We have a sphere with radius 2, and its radius increases by 1/8. What is the rate of change of the sphere's volume? \(\dfrac{4}{3}\pi r^3\) is the equation for volume, so should I just take the derivative of that and plug in 1/8? \[\dfrac{d}{dx}\dfrac{4}{3}\pi r^3 = 4\pi r^2\]\[4\pi (1/8)^2 = \dfrac{1}{16}\pi\]I don’t think that looks right... 
20181224, 17:45  #137 
"Curtis"
Feb 2005
Riverside, CA
1383_{16} Posts 
You're missing some key words from the problem, I believe; possibilities:
1. The radius is increasing at some *rate* (per unit time), in which case this is a related rates problem. If you haven't covered implicit derivatives yet, this isn't the case. 2. You're working on the section on differentials, in which case you should plug in the current radius for r and the change in radius for dr. Your solution used "d/dx", which doesn't make sense when there is no x in the equation nor the problem. You meant to write dV/dr, where V stands for volume. The idea used in differentials is that you can think of dV and dr as two separate quantities, rather than just a name for "derivative": dV as the change of volume, and dr as the change in radius. If you have an equation with dV/dr on one side and the derivative on the other side, you can multiply dr over to the right and plug in 1/8 for dr. Presto! A value for dV results, representing the change in volume that happens when r changes by 1/8. Note this value depends on the initial size of r (as it should!). 
20181225, 17:24  #138  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
Quote:
So we would have \(\dfrac{dV}{dr}\) on the left side, and the equation for volume of a sphere on the right, with dr being 1/8 and r being the original radius, 2? Which means dV is equal to 1/8 * 4/3 * pi * 2^3, which is \(\dfrac{4}{3}\pi\)... 

20181225, 19:25  #139  
"Curtis"
Feb 2005
Riverside, CA
1001110000011_{2} Posts 
Quote:
Yes, I believe your solution is correct. 

20190109, 01:20  #140 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}×3×37 Posts 
2 things I was thinking about, from last month's exam:
If I have \[\dfrac{x+1}{\sqrt{x2}}\]and rewrite it as \((x+1)(x2)^{\frac{1}{2}}\), then I can skip using the quotient rule and make it a bit simpler? I can't remember the problem exactly; maybe it had 1 or some other constant as the numerator... What does this notation mean? \[\dfrac{dx^2}{d^2y}\]That may not have been the exact placement of the exponents, but I remember that they were in alternating positions. Perhaps it means to take the second derivative... 
20190109, 07:43  #141  
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
10954_{10} Posts 
Quote:
https://en.wikipedia.org/wiki/Second_derivative 

20190109, 09:51  #142  
Dec 2012
The Netherlands
5·349 Posts 
Quote:
so it doesn't make very much difference. Use whichever you find easier. Yes! Thinking about the questions long after the exam is over is the sign of a true enthusiast! 

20190109, 17:18  #143 
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
25312_{8} Posts 

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