Recent content by Xishem

Ok. First, what is: \frac{d}{dx} (-x^3+3x)

Your derivative is incorrect. Specifically, this part: \frac{d}{dx}3x

[Edited out various confusions] ... I think I found the issue. Your definition of dV is inconsistent with your angle definitions. It should be dV=r^2 sin\phi\ dr\ d\theta\ d\phi

Everything seems ok to me. Why do you say that equals 0?

Multiplication is commutative, so your differential quantities can be multiplied and any order and will still be correct. You're missing some non-differential terms out front of your expression. Edit: Ninja'd

The derivative of a constant is 0.

Between lines 3 and 4 of your work you incorrectly simplified the first term like this: t^{\frac{1}{2}}(-2t)=-2\sqrt tOtherwise, I agree with your work.

Let's try this first: What is \frac{d}{dx}x^2 ?

You're using the power rule incorrectly also. The power rule is: \frac{d}{dx}cx^n=ncx^{n-1}Edit: Ah, it looks you are trying to do the product rule. Remember that if you have any constant times x or x divided by a constant, you can use the power rule instead of the product or quotient rules...

Instead of using the quotient rule, try writing the function as: \frac{1}{18}x^2 Now it is just a power rule derivative. Does that help? Not that the quotient rule doesn't work, though. From your answer, it looks like you're doing it incorrectly. Remember that the quotient rule is...

The object is originally headed on a vector of 270°. It changes course by -25° from 270°. Thus, the angle should be labeled from the 270° axis to the vector of the newest direction. You should be able to visually see that the angle you labeled in the OP is not 25°, as it is clearly obtuse.

If Earth did not rotate, one day would be the same as one year. One day is generally accepted as the time it takes for the Earth to perform a full rotation. If the Earth did not rotate, one full revolution around the sun would simulate a normal 24-hour day as far as the sun's motion relative to...