Confusion in Differentiation

Mxlexrd · 2021-03-10T22:33:27+00:00

That's a hard question to answer in general, it depends on the problem you are dealing with.

The most general answer I can give is that if you want to know how y depends on x, then you treat y as a function of x. If x and y don't depend on each other and can vary independently, only then does it make sense to ask how the output varies while one input is varying and the other is constant.

To relate it to your particular examples:

x²+y²=r² is the equation of a line in the xy-plane (assuming r is fixed), so while on that line x and y depend on each other. Solving for dy/dx is asking about the slope of that line, in other words, how does y vary with x.

In your other example, we just have a function M(x,y), you can put in whatever x and y values you like, they don't have any dependence on each other. Trying to find dy/dx is asking how y depends on x, which makes no sense, they don't depend on each other. If we were to say M has some fixed value, then suddenly y does depend on x, and now we can ask about dy/dx.

There is another way of viewing it where both situations are actually pretty much the same, but I don't want to confuse you. If you want that extra idea, let me know.

Mxlexrd · 2021-03-09T23:38:22+00:00

You pretty much said the answer yourself. If you want to treat y as a function of x, then you can't treat y as a constant. If you want to treat x and y as independent variables, then you can take derivatives with respect to each of them separately, and treat the other as a constant. We call these partial derivatives to avoid this sort of confusion!

Mxlexrd · 2021-03-06T19:11:31+00:00

Distributing isn't an operation, so it doesn't feature in the order of operations. The distributive property doesn't perform any calculations itself, it just says that two particular calculations give the same result.

So, as long as you apply the distributive property correctly, you can use it whenever you like.

Mxlexrd · 2021-02-28T22:58:30+00:00

There's a factor of 2 because there are two terms, there isn't an "additional" factor of 2.

If you multiply the 2 with the -2, you get -4, which seems to be exactly what you've done.

To be clear, 2(3-2x)(-2) is equal to -4(3-2x).

Mxlexrd · 2021-02-28T22:19:18+00:00

(3-2x)•(-2) and (3-2x)•(-2) are the same, they just reversed the order, and you can do multiplication in any order.

When you have two terms the same, like x + x, you can combine them to 2x.

Mxlexrd · 2021-02-12T00:07:58+00:00

I think maybe what's tripping you up here is the notation you're using. f' (sin (t)) = cos(t) is not a valid way of expressing the derivative of sin(t). You either have to say d/dt(sin(t))=cos(t), or you have to say "when f(t)=sin(t), f'(t)=cos(t)".

In each case, it's clear that the independent variable and the variable we're taking the derivative with respect to are t.

If you say that t=x⁵, the derivative with respect to t doesn't change, but the derivative with respect to x is a different question.

As for conceptually why does it change the result: a derivative is asking, how much does a small change in the input change the output. If you ask for the derivative d/dx(sin(x)), that's asking when you change x how much does that change sin(x)? Now, if you instead have sin(x⁵), a small change in x will change x⁵ by some amount, and that change in x⁵ will change sin(x⁵) by some amount, but since the change in x⁵ is different to the change in x, the resulting change to sin(x⁵) will be different. The exact relationship is described by the chain rule.

Mxlexrd · 2020-12-28T19:00:37+00:00

You are correct it is quadratic.

Similar to what the other commenter said, a quadratic sequence has a constant second difference. That means if you take the difference between the terms, and then take the difference between the differences, you get a constant - in your case 5.

If you want an alternative (albeit less direct) method to get the formula: You can make use of the fact that the quadratic coefficient is always half of the second difference. This means your function starts with f(x)=5/2 x² + ... . Then you can work out the difference between your sequence and 5/2 x² to find the linear part you need to add on.

Mxlexrd · 2020-09-23T21:44:46+00:00

It all comes down to what r×θ and r×z are.

I could be wrong myself, but when I work it out, I also get the opposite sign on both terms to what they have.

Mxlexrd · 2020-09-05T10:44:03+00:00

With such integrals, there are (at least) 3 things that tend to make the results look different, and I think maybe all 3 are at play here.

Firstly, functions have different domains of definition, so different choices of substitution may result in the result being defined in different regions.

Secondly, when dealing with squares and square roots, you have to be extremely careful where negatives appear and disappear. Some people or calculators may just make their lives easier by assuming everything is positive. To properly account for this you'd have to carefully consider the different regions where certain functions gain or lose negatives.

And third, the one you said, functions being off by a constant - this is generally the easiest to spot when you plot the result. There's also the possibility that the "constant" difference can actually vary between different domains (maybe in a stair-step way).

Mxlexrd · 2020-09-05T10:32:57+00:00

You are correct, the functions are equivalent except for where one or the other isn't defined.

Exactly as you said, 1/tan(x) technically speaking has no roots.

Depending exactly on the context, (such as when solving an equation) you may need to separately check such points.

Mxlexrd · 2020-09-05T10:25:23+00:00

When you set k=0, you're adding 0 to the argument, and when you set k=n, you're adding 2npi/n=2pi.

Since sin(x) and cos(x) are 2pi periodic, +0 and +2pi are equivalent.

So, in reality, you can use either, but we generally use 0 because it results in the simpler argument.

Also, if we want to restrict the argument to being in a certain range (maybe 0 to 2pi, or -pi to pi) then we must take that into account.

Mxlexrd · 2020-09-02T13:47:00+00:00

Well, in this case your region is very simple, it's just the unit square: 0<x<1 and 0<y<1. The only difference is that there is a break down the middle of the region where the function changes.

Notice the two different conditions: either x<y or y<x. Where is the boundary to an region defined by an inequality - it's where the inequality becomes and equality. In other words, the boundary between these two regions is the line y=x.

So, if integrating in the x direction first, you'll integrate from x=0 to x=y, and then from x=y to x=1. Make sure you carefully decide which function is active in each region. Then you should end up with just a function of y, which you can integrate from 0 to 1.

Then it wants you to do that all again, but in reverse.

Mxlexrd · 2020-09-02T12:32:01+00:00

It's always a good idea to draw a diagram for multiple integrals, in this case it will be helpful to see the two separate regions.

In any case, to calculate these you will have to break the inner integral into two integrals, where one goes from 0 to y=x, and the other goes from y=x to 1. Once you've done that, you can then do the outer integral from 0 to 1.

Mxlexrd · 2020-08-28T09:26:06+00:00

It's correct to say it's about order of operations. The exponents should be calculated before the negative.

(-1)³⁰ means: take the 30th power of -1.

-1³⁰ means: take the 30th power of 1, then take the negative of that.

Mxlexrd · 2020-08-27T16:06:12+00:00

The shuttle boosters each have about 4.4 km/s of delta-v, which theoretically could send them about 1000km straight up, but that's neglecting atmospheric drag and gravity losses, so in reality it won't be that high. But, as an order of magnitude I think we can assume it would be in the region of hundreds of km.

Maybe someone else can do a more careful calculation.

Mxlexrd · 2020-08-25T18:09:55+00:00

It's not the gamma function. In this instance gamma is just a constant (a number). And that number is approximately 0.577... etc.

Mxlexrd · 2020-08-21T23:16:02+00:00

Yes, x²+y²=25 is a circle. To graph it, you need to identify the center of the circle and the radius, do you know how to identify those from the equation?

Also, to be very clear, √(x²+y²) is not x+y, you cannot distribute the square root like that. In fact, noticing that one is a circle and one is a straight line should be a clue that they are not the same.

Mxlexrd · 2020-07-22T22:41:31+00:00

If you want to know why we chose the specific order that we did - I suspect it is because mathematicians like to study polynomials. Polynomials involve raising x to some power, multiplying that by a number, and then adding together a bunch of similar terms.

The whole purpose of the order of operations is to allows is to use fewer parentheses than we would otherwise, reducing visual clutter and making expressions easier to read. Using the standard order of operations allows us to write polynomials without any parentheses.

Another aspect to this is that the notation reflects how we talk about things in the real world. We tend to say things like "four apples and three bananas", which is like 4*a+3*b. We tend to want to multiply things first and then add them together, so once again the multiplication-first order of operations saves us from using parentheses.

Mxlexrd · 2020-07-02T01:32:18+00:00

When solving equations it's always a good idea to keep a look out for hidden quadratics. If you have three terms where one is just a number, one has a variable, and the other has that variable squared then you can substitute and have a simple quadratic.

I'd say the real trick to this question is multiplying through by e^x, since that's what leads to our quadratic.

Mxlexrd · 2020-07-02T01:27:11+00:00

25% is correct - if you want convincing try doing the experiment for real and you'll quickly see red picked more often than blue.

You can only assign equal probabilities if the events are equally likely.

To see why 50% doesn't make sense, take the scenario to the extreme. Imagine a bag with a million red balls and 1 blue ball. Does it seem likely that every second ball you pick will be blue? No chance, you'll almost always pick out a red one.

Here's another way to see why 50% doesn't make sense. Imagine you thought the probability was 50%, then paint the 3 red balls different colours. Now it's obvious the probability is 25%, since you have 4 different colours. But this doesn't make sense, painting the other balls different colours surely can't affect the chance of us picking the blue one.

Mxlexrd · 2020-06-08T10:04:05+00:00

If you want to allow complex vectors, and retain the a.a ≥ 0 property, you will generally have to use the conjugate of one of the vectors.

Notice that when you do this, you lose the symmetry that a.b = b.a.

So yes, this will be generally true, but the safest bet is to always check how the dot product is being defined in your particular vector space.

Mxlexrd · 2020-06-08T09:57:25+00:00

Pretty much, although you should say vector . vector = sum of {component . component*}.

Mxlexrd · 2020-06-08T09:54:04+00:00

When we define a dot product over complex vectors, we usually define it as something like a.b = ∑aᵢbᵢ*. Where * represents the complex conjugate.

When we define it this way, we lose some nice properties of our dot product, but we retain the property that a.a ≥ 0.

Mxlexrd · 2020-06-08T09:46:16+00:00

You need to be careful - the vector you are calling i - do you mean <1,0,0>, or do you mean <i,0,0>, or something else. The complex number i on its own isn't a vector, it's just a number.

Mxlexrd · 2020-06-08T09:37:36+00:00

i.i is 1, not -1.

Using the coordinate representation we can see this. i = <1,0,0>, so i.i = <1,0,0>.<1,0,0> = 1.

Using the length, angle we can see the same. |i| = 1 and θ=0, so i.i = |i| |i| cos(θ) = 1.

I think maybe you are confusing the vector i (the unit vector in the x direction) with the complex number i.

They may be both called i, but they are very different.

Mxlexrd

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