Ordinary differential Equation....I think
- Thread starter c4l3b
- Start date
\(\displaystyle f(r,s)=cos^{2}(3rs)+ln(r^{2})+5scos(rs)+s+6\)
This is not an exercise in DE's, but rather partial derivatives.
They want you to differentiate w.r.t r and s and so on.
When finding, say, df/dr, treat s as a constant and differentiate w.r.t r.
Take the derivative of each term. They aren't bad.
Here is the first one:
\(\displaystyle \frac{df}{dr}=-6s\cdot sin(3rs)cos(3rs)-5s^{2}sin(rs)+\frac{2}{r}\)
See?. The derivative of \(\displaystyle cos^{2}(3rs)=-6s\cdot sin(3rs)cos(3rs)\) because we held s as a constant and just took the derivative as usual. If we find df/ds, then hold r as a constant.
To find \(\displaystyle \frac{d^{2}f}{dr^{2}}\), find the derivative again.
This is not an exercise in DE's, but rather partial derivatives.
They want you to differentiate w.r.t r and s and so on.
When finding, say, df/dr, treat s as a constant and differentiate w.r.t r.
Take the derivative of each term. They aren't bad.
Here is the first one:
\(\displaystyle \frac{df}{dr}=-6s\cdot sin(3rs)cos(3rs)-5s^{2}sin(rs)+\frac{2}{r}\)
See?. The derivative of \(\displaystyle cos^{2}(3rs)=-6s\cdot sin(3rs)cos(3rs)\) because we held s as a constant and just took the derivative as usual. If we find df/ds, then hold r as a constant.
To find \(\displaystyle \frac{d^{2}f}{dr^{2}}\), find the derivative again.
galactus said:
Galactus,
Thank you very much that has really helped me.
But I have noticed you have differentiated up to \(\displaystyle \frac{2}{r}\) is that completely finished or do you just keep on calculating to \(\displaystyle +5scos(rs)+s+6\), in other words do you calculate whole equation for each derivative?
Would you mind just showing the calculations for d^2f/dr^2, d^2f/ds^2, d^2f/drds?
Thanks
Yes, that one is complete. The \(\displaystyle \frac{2}{r}\) is the derivative of \(\displaystyle ln(r^{2})=2ln(r)\)
Since the derivative of ln(r) is 1/r, then we multiply by 2 and have it.
Instead of me doing those, how about you do them and let me know what you get. Then, I can help you along with any mistakes and what not. OK
Since the derivative of ln(r) is 1/r, then we multiply by 2 and have it.
Instead of me doing those, how about you do them and let me know what you get. Then, I can help you along with any mistakes and what not. OK
galactus said:
Ok,
so for df/ds is
holding r as constant
\(\displaystyle \frac{df}{ds}=-6r\cdot sin(3rs)cos(3rs)-5s^{2}sin(rs)+\frac{2}{r}\)
is that right?
And going back to the equation we started with, I believe you cant derive s + 6, so you just leave those?
Could you just show me how to do \(\displaystyle \frac{d^{2}f}{dr^{2}}\)?
In df/dr, s and 6 were constants, so they're derivatives were 0.
In df/ds, the derivative of s is 1.
Here is df/ds. Remember, in this case, r is a constant, so \(\displaystyle ln(r^{2})\) is a constant and its derivative is 0.
\(\displaystyle \frac{df}{ds}=-6r\cdot sin(3rs)cos(3rs)+5cos(rs)-5sr\cdot sin(rs)+1\)
I relaize these things can be tedious when you first start doing them. Hang in there. It will get easier. Most of the time, one forgets to hold one of the variables as constant.
In df/ds, the derivative of s is 1.
Here is df/ds. Remember, in this case, r is a constant, so \(\displaystyle ln(r^{2})\) is a constant and its derivative is 0.
\(\displaystyle \frac{df}{ds}=-6r\cdot sin(3rs)cos(3rs)+5cos(rs)-5sr\cdot sin(rs)+1\)
I relaize these things can be tedious when you first start doing them. Hang in there. It will get easier. Most of the time, one forgets to hold one of the variables as constant.