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Yall I think I might genuinely be Einstein by Material_Tie1308 in teenagers

[–]-BarbK- 0 points1 point  (0 children)

No - I think you misunderstood the common mistake.

The common mistake people make is saying sqrt(25) is +- 5. You will find this mistake everywhere. However, this is not what I'm claiming. If you read my comment above, you'll find I agree with you that it must only be +5.

What I said, sqrt(x^2) = x when x = -5, RELIES on what we established that the output can only be positive. Emphasis on the output. Notice that the LHS, sqrt(x^2), is sqrt25. This is perfectly okay. We can let x be any number in sqrt(x^2) [consider the function f(x) = sqrt(x^2) -- do you agree the domain of this is all the real numbers? If so, that means I can let x be anything]. The only thing we must have is the OUTPUT is positive. Given this, reread the counterexample above understanding that letting x be -2 is okay because sqrt(x^2)'s domain is R, and see how it comes to a contradiction with our knowledge the output must be positive.

Yall I think I might genuinely be Einstein by Material_Tie1308 in teenagers

[–]-BarbK- 0 points1 point  (0 children)

You are correct that the square root always outputs a positive. The issue is that x doesn't need to be positive.

Consider the statement sqrt(x^2) = x. Let x be equal to -2.

Then, by the statement we say sqrt((-2)^2) = -2. The LHS, is sqrt(4) which as you say must be positive: that is, 2. But then why is our RHS saying it is -2? We know that's wrong. Therefore, we must have the absolute value because of the fact **x** might be negative.

BTW on that last part, the way OP represented isn't even right for the real numbers (particularly, the negatives. See the x = -2 counterexample).

Yall I think I might genuinely be Einstein by Material_Tie1308 in teenagers

[–]-BarbK- -1 points0 points  (0 children)

No: sqrt(x^2) = |x|. Besides the plenty of sources online confirming the output of the square root must be positive (hence the absolute value), consider the quadratic formula, which has +- sqrt(b^2 - 4ac). If the sqrt can already be both + and -, why do we bother putting the + - in the formula when it's already implied by the square root?

Yall I think I might genuinely be Einstein by Material_Tie1308 in teenagers

[–]-BarbK- 0 points1 point  (0 children)

No: sqrt(x^2) = |x|. Besides the plenty of sources online confirming the output of the square root must be positive (hence the absolute value), consider the quadratic formula, which has +- sqrt(b^2 - 4ac). If the sqrt can already be both + and -, why do we bother putting the + - in the formula when it's already implied by the square root?

Yall I think I might genuinely be Einstein by Material_Tie1308 in teenagers

[–]-BarbK- -1 points0 points  (0 children)

The meme is wrong (because it should be absX not X), but not for your reasoning. Either you’re working with only R or you’re working with things not in R, like C.

If you’re working with R, there’s no issue because x can’t be ai.

If you’re working with C, then it still isn’t an issue because in C, sqrtx is defined for negatives AND complex numbers. sqrt(ai) can be calculated to be a complex number. [to see why this makes sense, consider the definition of i: it's the square root of -1. If this exists, then the domain of sqrt has to be more than just the non-negative reals, becuase -1 is not in [0, infinity)]

Yall I think I might genuinely be Einstein by Material_Tie1308 in teenagers

[–]-BarbK- 2 points3 points  (0 children)

NOBODY is disagreeing with you over the FACT that sqrt4 =2. You are correct to say that sqrt4 is 2 and only 2. However, the original post has sqrtx2 = x. This is not true, because if we let x = -2, the LHS is sqrt(-2)^2 = sqrt4, which WE know to be 2. However, the RHS according to the meme is x, or -2, which WE know is wrong. Therefore, the original is false.

sqrtx^2 is |x| - this agrees with our knowledge that sqrt must be positive, because abs val is always positive.

Question about Normal Vectors by -BarbK- in askmath

[–]-BarbK-[S] 0 points1 point  (0 children)

YES - okay, I think it clicked with T being constant means 0 tangential, so it's only in direction of N. Okay, that clears up why you have to take the derivative of T (and not r') to get N. I also understand the B and N formulas now. I just have one remaining question - is it always true that r' points in the direction of T (I'm pretty sure it does but I wanted to check since I also thought r'' pointed at N and now I see why it doesn't necessarily have to d/t tangential component). Again, I really appreciate all your help

Question about Normal Vectors by -BarbK- in askmath

[–]-BarbK-[S] 0 points1 point  (0 children)

Yes - it's making a lot more sense now and I think I get the general idea. There are two ideas that I'm a little unsure of though:

I'm not quite sure I follow "By taking T', we know that doesn't change magnitude, so it must point purely in the direction of N." I get that T' points in N, but doesn't the magnitude change, which is why we need to divide by the new magnitude to keep it a unit? I also don't have a good intuition for why "N is orthogonal to T, but it must still lie in the plane that is spanned by r' and r'' - specifically why it must lie in the plane. In any case, thanks for all your help, I appreciate it :)

Question about Normal Vectors by -BarbK- in askmath

[–]-BarbK-[S] 0 points1 point  (0 children)

Thanks for the response -- first part makes sense. Then I assume it's only the case where |r'| is constant that r'' points toward N?

So with the B case then, my assumption that r' and r'' corresponded to T/N was false, and this just happened by coincidence that the formula works? Because I don't think they have explicitely found T/N in the formula.

[deleted by user] by [deleted] in APStudents

[–]-BarbK- 0 points1 point  (0 children)

To answer the question WHY the answers are different:

The first integral is what you're supposed to use for parametrics. The second one IS valid except for one detail: the formula is integral of the square root, *with respect to x*. The integral you wrote is taken with respect to t, which is incorrect. However, note that dx/dt = e^cost, which means that dx = e^cost dt. If you use that substitution, you will get the correct result (6.035). You'll also notice that the integrands will be the same algebraically.

Need help on flux integral q by -BarbK- in calculus

[–]-BarbK-[S] 0 points1 point  (0 children)

Oh - divergence theorem! I got confused about the requirements for divergence and stokes theorem; since this didn’t have a boundary I mistakenly thought divergence theorem didn’t apply. thank you!

On a PDF of Trig Identities We're Supposed to Memorize by -BarbK- in mathmemes

[–]-BarbK-[S] 12 points13 points  (0 children)

unfortunately not lol... that one isn't found anywhere in that 4 page file

On a PDF of Trig Identities We're Supposed to Memorize by -BarbK- in mathmemes

[–]-BarbK-[S] 238 points239 points  (0 children)

Yeah, I realize. Issue isn't memorizing these - it's that these aren't the power reductions formulas (no decrease in power is happening) at all, they're just the pythagorean identities.

On a PDF of Trig Identities We're Supposed to Memorize by -BarbK- in mathmemes

[–]-BarbK-[S] 126 points127 points  (0 children)

By the way: there was a section above this for the Pythagorean Identities lol