# TD

Senior Members

52

## Community Reputation

10 Neutral

• Rank
Meson
• Birthday 06/20/1986

## Profile Information

• Location
Brussels
• College Major/Degree
Engineering Student
• Favorite Area of Science
Mathematics
1. ## Dedicated Mathematician

They're asking for the possible values of "5x+2", not of x. So when you say 0 is a possibility, this doesn't mean "x = 0" but "5x+2 = 0". C is the correct answer.
2. ## integral question

The problem isn't the 4x, but the square! Since you can easily integrate sin(x), it's no problem to integrate sin(ax) with a an arbitrary constant either! To get rid of the square, use another trig identity: $\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = 1 - 2\sin ^2 x \Leftrightarrow \sin ^2 x = \frac{{1 - \cos \left( {2x} \right)}}{2}$ Now you lost the square, went to a cosine and doubled the angle - but that shouldn't be a problem
3. ## Optimization

I can't check your answer since the question isn't 100% clear to me. Is B located 6 miles under the road, vertically? So the distance between B and the road is 6. And the 10 miles, is that the (straight) distance between A and B or between A and the point on the road which is above B?
4. ## non intuitive way of solving cos(X)=cos(x+pi/2)

Ok Well in general for equations like this, we have that $\cos \alpha = \cos \beta \Leftrightarrow \alpha = \beta + 2k\pi \vee \alpha = - \beta + 2k\pi ,\forall k \in \mathbb{Z}$ I don't consider this as "intuitive" since this holds in general and is a perfect algebraic solution - but perhaps the topic starter can clarify what he meant.

6. ## non intuitive way of solving cos(X)=cos(x+pi/2)

What do you mean with non-intuive way? Two cosines are the same when either the arguments are the same or opposites, with 2k*pi of course.
7. ## Optimization

You're on the right track. With your drawing, split the problem in 2 parts using an intermediate point P. Express the distance as a function of the distance to P by using the pyth. theorem on the triangle, compute the time necessary to do both parts with the given speeds and then add those two parts.
8. ## Help wid limit Question

There's no need to apologize, but do you see what your mistake was? That's quite fundamental, in general (a+b)^n isn't a^n + b^n.
9. ## Help wid limit Question

The answer is indeed 1/27 but be careful, b = 3, not -3 in this case (you substituted correctly though).
10. ## Help wid limit Question

I'll go into a bit more detail. We would like to get rid off that cube root to simplify the numerator. For square roots, we can use the fact that $a^2 - b^2 = \left( {a - b} \right)\left( {a + b} \right)$ but for cube roots, that won't help. As I said, we'll be using the identity $\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) = a^3 - b^3$. We consider the current numerator as the factor (a-b) with of course $a = \left( {x + 27} \right)^{1/3}$ and $b = 3$. Now we multiply numerator and denominator with the same factor, namely the second one of our identity, so (a²+ab+b²) with our a and b. After doing that, instead of cancelling these equal factors (then we wouldn't have done a thing...) we can simplify the numerator since the expression there is now equal to a³-b³ according to our identity. But with our a and b, that becomes $a^3 - b^3 \Rightarrow \left( {\left( {x + 27} \right)^{1/3} } \right)^3 - 3^3 = x + 27 - 27 = x$ and the cube root is gone, just as we wished. The only thing that's left is an x, but that can be cancelled out with the x in the denominator leaving only our added expression in the denominator. It's now possible to simply fill in x = 0 in our limit to find the value.
11. ## Help wid limit Question

Eeks! $\left( {a + b} \right)^{1/3} \ne a^{1/3} + b^{1/3}$ !!!
12. ## Help wid limit Question

Use the fact that $\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) = a^3 - b^3$ and consider the nominator as the factor (a-b) in that expression. Then find out what a and b are and multiply nominator & denominator by the second factor in that expression. The nominator now simplifies to a³-b³ and you'll be able to cancel out an x which will get rid off the indeterminate form. Then it's just filling in I basically said it all, now it's up to you to do the numbers!
13. ## Nothing multiplied by infinity equals a finite number

That's true but I don't see the relevance of your complex example. Also in the reals, a limit isn't defined if it depends on the way you approach it (meaning: it has to be the same, coming from the left or from the right). In that way, the limit for x going to 0 of 1/x isn't defined because depending on coming from the left or right, you get - or + infinity.
14. ## Nothing multiplied by infinity equals a finite number

Correct, with that nuance that it could be negative infinity as well of course -depending on the sign of the numerator and the way of approaching 0.
15. ## Nothing multiplied by infinity equals a finite number

That's not correct. 0*(a finite number) is 0 but 0*inf is an indeterminate form, just as 0/0, inf/inf, inf-inf, ...
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