You are not logged in.
Good point. I think only Anastasia can clear that out ...
Hello fellow Mathematicians & non-Mathematicians,
As I'm new to the forum, I thought to introduce myself. Obviously, I'm majoring in chemistry , and I'm still in my 1st semster. I'm currently taking Calculus III & Analytical geometry as well as Statistics & probability.
There are two more Math courses to go during the next 2 semesters, so I'm counting on you geniuses.
Yes you're right , I apologize for my mistake, for x=0, y=1. Isn't a vertex considered like an intersection point btw?
Greetings Flowers4Carlos,
Thank you for your help. I believe there's no mistake in your calculation, which is pretty good and simple. I shall double check the answer anyhow, thanks again.
you mean finding the roots of this quad eq ?
General formula :
ax^2 + bx + c = 0
First find Delta Δ = b^2 - 4ac
x1 = ( -b + √Δ ) / 2a
x2 = ( -b - √Δ ) / 2a
If b is even , you can take delta = b^2 - ac , if i remember correctly, and instead of 2a, u use a in the denominator.
Anyway the roots should be :
x1 = - 0.29
x2 = -1.7
Concerning the intercepts, just equate x and y to 0 for a general equation. In this case, this is a parabola, so the roots found are x intercept and the y intercept is 0.
No problem mate, I'll try out different methods. Thanks anyway.
Hello mathsyperson ,
thank you for your help, but you obviously didnt notice the / ( division ). The exp is (1 + cosθ) / (1 - cosθ). This is why I changed it to TanB/2
Greetings mates,
If anyone can help me solve the following integral using the change of variable ჯ = cosმ
The ^ means to the power, so 2Cos^2მ is 2 multiplied by the squared of Cosმ ( just to avoid confusion ).
The given integral is:
I = ∫[(1+ჯ)/(1-ჯ)]^½dჯ
I already worked it out using other methods like taking t^2 = (1+ჯ)/(1-ჯ) and then solving in terms of t. I found out the solution. However, the method of change of variable using Cosმ is not working. For starters,
I = ∫[(1+cosმ)/(1-cosმ)]^½dcosმ = - ∫[(2Cos^2B)/(2Sin^2B)]^½ sinმdმ where B = მ/2
This implies, I = - ∫[1/|Tan^2B|] sinმdმ
--> I = - ∫ (sinმdმ / Tan^2B )
How to cotinue from here? I've tried many methods using integrations by parts, by nothing worked out.
Thanks in advance.