If you are looking to integrate over the whole real line, and are familiar with statistics, another approach is to notice that, apart from a constant, this is similar to finding the expectation of x^2 where x is a normally distributed random variable with mean 0 and variance a^2. Using the...
I can only imagine you are getting intimidated by the sin(x) and cos(x) functions. Try writing y = sin(x) and x = cos(x) then simplify. Hopefully going through this exercise will be helpful.
The slope won't change anything because, in the absence of friction, PE + KE = constant (as you have rightly used in your calculation). So it's the magitude of the change in PE that affects your KE at the end, not how you get there.
It's a property a "conservative field" which you may want to...
Both questions are to do with conservation of energy (surprise!).
Try not to get caught up in thinking it's more complex than this - Ask yourself what kind of energy they have at the start of the experiment and what they have at the point of measurement (no friction so no losses to worry...
I like StatusXs explanation, but if you want to work with KE, here is the most explcit way to deal with what you have been given.
Call the heavier truck 1, and the lighter truck 2. Since KE is the same
\frac{1}{2}m_{(1)}V_{(1)}^{2} = \frac{1}{2}m_{(2)}V_{(2)}^{2}
Since m_{(1)} >...
The answer should be apparent from the information you write at the start. In general, if you weren't already given what the sum is, you might was to consider looking at the following to see if provides any hints.
S(4) - S(3)
S(3) - S(2) etc.
See if any pattern arises which allows you to...
Think of a volume of revolution formed by a general line y = mx. What sort of shape does this yield and how is it related to the shape you are given?
Thinking about this will also provide a good check as to whether your answer is correct as there is a nice formula for its volume.
I'm not sure I understand the question fully but here are my thoughts:
- It depends on whether k < 1 or k >= 1 .
- Since you say it's k times greater, i'll assume k > 1 but this creates a less interesting problem than the other case.
You can reason without the help of any formula that...
There are two ways to set up this problem:
1. Use the usual equations of motion for body 1 (the heavier mass) and 2 (the lighter mass).
S(1) = 70 -10t - 0.5gt^2
S(2) = 60 - 0.5gt^2
Then solve for the time when their positions are the same.
2. Since the acceleration of both...
As a quick tip, if you want to check if you have performed manipulations like this correctly, put some numbers in to make sure that both sides of the equation are still the same.
I would say that, if showing how you get to the equation is part of the coursework then it's likely to be a useful tool for your course. (Not to mention that basic algebra isn't a bad thing to know generally!)
It's probably best for you in the long run that you have a go and show us where you...
I guess it could also have been:
\frac{dy}{dx}=-y+a(x)y^{2}+b(x)
but given the requirements of part 2, I think you are right, it should be:
\frac{dy}{dx}=-y^{2}+a(x)y+b(x)
Yoyo - the problem you have been given is very explicit about what is required. If you show a little of what...