00:01
So this question first wants us to estimate the derivative at various points using the symmetric difference quotient.
00:10
So it says that f prime of x is approximately equal to f of x plus a minus f of x minus a all over 2a.
00:28
And in this case it wants a to be 10.
00:32
So p prime of our first point 303 is approximately p of 10 plus that point minus p of that point minus 10, all over two times the width, then we can do this again.
01:08
P of point plus 10, minus p of the point minus 10.
01:24
All over 20.
01:28
This first one works out to be 0 .00265, and this second one works out to be 0 .004145.
01:46
Then it wants us to keep doing this.
01:49
So p prime of 3 to 3 is approximately equal to again p of the point plus 10 minus p of the point minus 10 all over 2 times 10 this one works out to be 0 .006295.
02:47
And this one works out to be .009305.
02:58
And then our last approximation is p prime of 343, which is p of the point plus 10 minus p of the point minus 10 all over 20.
03:19
And that works out to be 013435.
03:26
So now that we have our derivatives estimated, it wants us to see if dpdt is proportional to some constant k times p over t squared...