hi Zeeshan 01
I did this integral a different way.
So my first thought when I saw your answer was ???
So I plotted the two functions together on a graph. The second plot fitted exactly over the first. To show this more clearly I have offset my graph by 0.3 along the axis so you can see both graphs.
Then I checked to see if I could prove they are the same by using identity methods:
So we have the same answer.
hi Zeeshan 01
What do you mean by rationalise here?
This is called rationalisation because the irrational denominator has been made into a rational.
But cosine(x) isn't an irrational. ??
To do the integral you can make use of
This makes a function that is directly integrable.
and find their velocities before and after the collision
There is not enough information to say what these are absolutely. eg cannot say p = 5 m/s.
But you can plug in the zero components into the momentum equations to say what the after-velocities are, in terms of p and x. That's as far as this problem will go without more information, I think.
Where did this question come from?
Have you looked at this page?
There's good news and bad news.. I'll deal with the bad first.
Each sphere has a line of direction before the impact and these meet at a point. Two lines meeting at a point define a plane so the action must take place in a 2-D plane. That means you cannot just use a single variable for the velocities. You'll have to have 2 components for each before the impact and 2 after. That seems like a lot of unknowns.
But now for the good news.
I'll take the x axis as along the line of centres. So for the 4Kg before impact I'll use components Ux and Uy.
And Vx and Vy for the 4 Kg afterwards.
And I'll use Wx and Wy for the 2 Kg before. I don't even need anything for afterwards because we're told it is brought to rest.
So now for some equations:
conservation of momentum in the x direction:..................1
I think the experimental law only applies along the line of centres as there is no 'bounce' in the tangential direction. If that seems strange consider a sphere bouncing on a surface. If we drop the ball onto the surface, it bounces according to the coefficient of restitution. But if you roll the sphere across the surface there's no bounce at all.
When the spheres touch they share a common tangential plane at the point of impact The radius of each at that point is at right angles to the tangential plane. So the line of centres is the x axis. So apply the experimental law to the x components:......................3
I'm a bit unsure about this next bit but it does yield the required result and it won't if it isn't true so here goes:
Perpendicular to the line of centres I think the relative velocities are equal so:..............................4
1 + 3 tell us about Ux and 2 + 4 tell us about Wy. That will give you the first required result.
Off to do some more research on this now.
hi Zeeshan 01
When you differentiate y wrt x you are finding the gradient function of y . That is what differentiation is.
have a look at these two graphs.
In the first, pick any point on the graph and imagine a line drawn at that point with the same gradient as the curve. We say the line is a tangent to the curve at that point. Make a note of the gradient at that x coordinate and consider on a new graph the point (x, gradient). Do that for all of the points on the curve and you get a new curve which is called the gradient function of y.
For y = x^2 let's think about what this gradient function is like. when x is positive, the gradient is positive. Because the curve y = x^2 is symmetrical the gradient at a negative value of x will be the same as at the positive value of x, except it will now be negative. For example, at x = 2 there will be a certain gradient. At x = -2 the line slopes exactly the same amount but downwards from left to right. So if the gradient at x = 2 is m, then the gradient at x = -2 will be -m. At x = 0 we can see that the gradient is also 0.
It is possible to show that the gradient function is 2x so we can write dy/dx = 2x. But even without knowing the theory for this you should be able to see that 2x has the right properties. It is zero at x=0; at x = 3 it is 6; and at x = -3 it is -6.
Now what about the gradient function for a^2. The graph plotter required that I choose a value for a, so I chose a = 3. The graph for y = a^2 is a straight horizontal line. The gradient is zero at every point. So its gradient function is zero at every point. So d(a^2)/dx = 0.
If 'a' varies as x varies then this would not be so. But, in the context of your question, it seemed reasonable to assume that a is a constant. If that is not so then d(a^2)/dx = 2a.(da/dx). In other words, it would have a similar form to the gradient function for y^2.
Why differentiate wrt x? Because you wanted dy/dx. The dx tells us what you have to differentiate with respect to.
Hope that explanation helps. So have you got a final answer for y2 ?
Sorry to hear about your flu. Hope you're soon feeling better.
I find the key to many geometry problems is a good diagram. Based on those angles your is a bit lop-sided. Is this better?
It shouldn't take long to chase around those angles in right angled triangles. If my diagram isn't correct please would you specify the angles like this DAC = 90 etc.
edit: hhmmm. Mine cannot be right as DM = 10 and MA = 4. Please help correct mine.
hi Zeeshan 01
Excellent! That tells me two things. (1) You know about 'implicit' differentiation (2) That's the way you'll want to do the next step.
But first let's just recap how you got there.
So x and y are variables and 'a' is a constant.
When we differentiate that equation each part must be differentiated with respect to (wrt) x.
x^2 is easy enough. That becomes 2x.
But what about y^2. This is not even a function of x. You have to make use of the chain rule:
In this case z = y^2
So we differentiate wrt y and then multiply by dy/dx. So we get
Now we must differentiate the a^2. But 'a' is a constant. It's gradient function is zero whatever x is. So
Putting all this together and re-arranging you end up with
Now to differentiate this.
The right hand side is a 'quotient' or fraction so we must apply the rule for that:
If you have not met this rule before, please ask about it.
In your example p = y and q = x so we get
You can do a little simplifying from here.
Welcome to the forum.
You have come to the right place. When you encounter a bit of algebra that you cannot understand, post the details here, and we will help you. I recommend you join as a member.
How are you with arithmetic? This might seem an odd thing to ask but it is relevant. The rules of algebra are the rules of arithmetic. When I'm teaching someone an algebraic topic I usually start by asking "How would we do this with numbers?"
There's also a wealth of pages plus practice questions here:
You need to eliminate one of the unknowns. For this example it is easiest to replace the c in the second equation with the value of c from the first like this:
s + (2s - 650) = 1804
From there it is just a single unknown equation. After you find s you should be able to use the first equation to find c.