RESOLVED! Struggling with an urgent calculus assignment for a degree

[joebloggs7654]

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!

[krtxmrtz]

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

[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

[krtxmrtz]

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.

[krtxmrtz]

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".

[krtxmrtz]

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...

[joebloggs7654]

Huge thanks krtxmrtz that has been extremely helpful to me. Resolved!

Kind regards + Many many thanks

Joseph

[krtxmrtz]

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!

[Mattywoo2]

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.

[Mattywoo2]

Does that link work? It should

http://geocities.com/matthewfriend2001/vbthing.pdf

[krtxmrtz]

Does that link work? It should

http://geocities.com/matthewfriend2001/vbthing.pdf
It works.

I used this elaborate method as I not too familiar with hyperbolic functions... :)

[Mattywoo2]

Oh poo, there would be middle terms when you square sech(x) + C
Never mind, you get the idea.

[Mattywoo2]

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