RESOLVED! Struggling with an urgent calculus assignment for a degree
I have a maths assignment to complete before Monday, and I'm really stuggling. Its to do with differentiation and integration. :(
The pdf document is here
www.geocities.com/joebloggs7654/this.pdf
and I would be very grateful if someone could help me out with question 6.
Many thanks
Joseph
Update: Resolved!
Re: Struggling with an urgent calculus assignment for a degree
Part a is not too difficult.
dy/dx = sin(3x) / [2 + cos(3x)]
Using I meaning "integral":
y = I(dy) = I{sin(3x)dx / [2 + cos(3x)]}
If you call z = cos(3x) then,
-3sin(3x)dx = dz
sin(3x)dx = -dz/3
cos(3x) = Sqrt[1 - sin2(3x)] = Sqrt(1 - z2)
dx = -dz/[3*Sqrt(1 - z2)]
Substituting,
y = I{-dz/[3*Sqrt(1 - z2)]} = -log(2 + z)/3 + K
where log is the base-e logarithm and K is an integration constant to be determined from the initial conditions:
(y, x) = (4, 0)
Injecting these values into the above equation:
4 = -log(3)/3 + K
K = 4 + log(3)/3
Thus:
y = -log(2 + z)/3 + 4 + log(3)/3 = 4 + log[3 / (2 + z)]/3
and finally, replacing z by its value:
y = 4 + log[3 / (2 + cos(3x)]/3
Re: Struggling with an urgent calculus assignment for a degree
Many thanks krtxmrtz, that is really helpful, as I can see exactly where you're coming from on this one, and you have helped me to understand the concept.
Joseph
Re: Struggling with an urgent calculus assignment for a degree
Quote:
Originally Posted by joebloggs7654
Many thanks krtxmrtz, that is really helpful, as I can see exactly where you're coming from on this one, and you have helped me to understand the concept.
Joseph
You're welcome. By the way, a "dx" was missing in line 4 of my previous post. I've just placed it there.
Re: Struggling with an urgent calculus assignment for a degree
Part b is a little more cumbersome as it requires 2 substitutions. First let's work on b-i.
dy/dx = 4y1/2(e-x - ex) / (ex + e-x)2
This can be rearranged as:
dy/2y1/2 = 2(e-x - ex) / (ex + e-x)2
The left side is immediate:
Sqrt(y) = 2(e-x - ex) / (ex + e-x)2
y = 4(e-x - ex)2 / (ex + e-x)4
and I imagine this is what they mean by "the solution in implicit form".
Re: Struggling with an urgent calculus assignment for a degree
Part b-ii:
dy/2y1/2 = 2(e-x - ex) / (ex + e-x)2 = R
Let's integrate the right hand side that I've tagged "R". This is my first substitution:
z = ex
Hence, dx = dz/z
I(R) = I[2(1/z - z)dz / (z + 1/z)2] = 2I[(1 - z2) / (1 + z2)]
My next substitution is:
z = tan(q) -> dz = dq/cos2q
Some algebra leads to
sin q = z/Sqrt(1 + z2)
cos q = 1/Sqrt(1 + z2)
which will be required later.
The integral becomes:
I(R) = 2I[(1 - tan2q)(dq/cos2q) / (1/cos2q)2] = (...after some manipulation...) = 2I(cos2q - sin2q)dq = 2I(cos(2q)] = sin(2q) = 2sin(q)cos(q)
Back substituting,
I(R) = [2z/Sqrt(1 + z2)] * [1/Sqrt(1 + z2)] = 2z / (1 + z2) = 2ex / (1 + e2x) = 2 / (ex + e-x) = 1/Ch(x)
i.e., the result is the inverse of the hyperbolic cosinus of x.
Finally,
Sqrt(y) = I(R)
therefore,
y = 1 / Ch2(x) + K
where I've just added the integration constant K.
I hope you can handle the other 2 parts...
Re: Struggling with an urgent calculus assignment for a degree
Huge thanks krtxmrtz that has been extremely helpful to me. Resolved!
Kind regards + Many many thanks
Joseph
Re: Struggling with an urgent calculus assignment for a degree
Quote:
Originally Posted by joebloggs7654
Huge thanks krtxmrtz that has been extremely helpful to me. Resolved!
Kind regards + Many many thanks
Joseph
You're welcome again, but I'd appreciate it if you would rate this post favourably!
Re: RESOLVED! Struggling with an urgent calculus assignment for a degree
Quote:
Originally Posted by krtxmrtz
Part b is a little more cumbersome as it requires 2 substitutions. First let's work on b-i.
Hi there.
This intergral requires no substitutions per se, just simplification using the hyperbolic functions. See my solution at http://geocities.com/matthewfriend2001/vbthing.pdf .
It was a nice solution though.
Re: RESOLVED! Struggling with an urgent calculus assignment for a degree
Re: RESOLVED! Struggling with an urgent calculus assignment for a degree
Quote:
Originally Posted by Mattywoo2
It works.
I used this elaborate method as I not too familiar with hyperbolic functions... :)
Re: RESOLVED! Struggling with an urgent calculus assignment for a degree
Oh poo, there would be middle terms when you square sech(x) + C
Never mind, you get the idea.
Re: RESOLVED! Struggling with an urgent calculus assignment for a degree
In fact, I got both bits wrong. When I separated the vaiables I didn;t take the constant with me. It should read
y^(1/2) = sec(x) + C (NOT 4y^(1/2)...)
So
y = sec²(x) + 2Csec(x) + C²
Matt