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Old 2018-12-22, 18:13   #133
jvang
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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.
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Old 2018-12-23, 09:34   #134
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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.
Try using the chain rule instead!
\[ \frac{d}{dx}(\sin x)^3=3(\sin x)^2\frac{d}{dx}\sin x=3(\sin x)^2\cos x. \]
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Old 2018-12-23, 22:47   #135
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Try using the chain rule instead!
\[ \frac{d}{dx}(\sin x)^3=3(\sin x)^2\frac{d}{dx}\sin x=3(\sin x)^2\cos x. \]
Oh, yeah. Duh!
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Old 2018-12-24, 16:36   #136
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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...
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Old 2018-12-24, 17:45   #137
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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!).
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Old 2018-12-25, 17:24   #138
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Quote:
Originally Posted by VBCurtis View Post
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!).
So the derivative notation is shorthand for a difference quotient sort of thing? That’s helpful to know...

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 \(\dfrac{4}{3}\pi\)...
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Old 2018-12-25, 19:25   #139
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So the derivative notation is shorthand for a difference quotient sort of thing? That’s helpful to know...

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 \(\dfrac{4}{3}\pi\)...
Consider what the definition of derivative is! You spent days taking limits of difference quotients; that's the foundation of everything you're doing with derivatives.

Yes, I believe your solution is correct.
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Old 2019-01-09, 01:20   #140
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2 things I was thinking about, from last month's exam:

If I have \[\dfrac{x+1}{\sqrt{x-2}}\]and rewrite it as \((x+1)(x-2)^{-\frac{1}{2}}\), 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...
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Old 2019-01-09, 07:43   #141
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Quote:
Originally Posted by jvang View Post
2 things I was thinking about, from last month's exam:

If I have \[\dfrac{x+1}{\sqrt{x-2}}\]and rewrite it as \((x+1)(x-2)^{-\frac{1}{2}}\), 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...
You don't need to ask us your second question.
https://en.wikipedia.org/wiki/Second_derivative
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Old 2019-01-09, 09:51   #142
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Quote:
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If I have \[\dfrac{x+1}{\sqrt{x-2}}\]and rewrite it as \((x+1)(x-2)^{-\frac{1}{2}}\), then I can skip using the quotient rule and make it a bit simpler?
Yes, you can do that, but then you need the product rule instead of the quotient rule,
so it doesn't make very much difference. Use whichever you find easier.

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2 things I was thinking about, from last month's exam:
Yes! Thinking about the questions long after the exam is over is the sign of a true enthusiast!
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Old 2019-01-09, 17:18   #143
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Yes, you can do that, but then you need the product rule instead of the quotient rule,
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!
Seconded!
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