00:01
Okay, we're given that the focal length f is related to an object's distance p and the image distance q by this formula, 1 over f equals 1 over p plus 1 over q.
00:12
Find the instantaneous rate of change of q with respect to p if f is constant.
00:19
Okay, instantaneous rate of change.
00:22
That means take the derivative of q.
00:27
So q is taking the place of y or f of x with respect to p.
00:32
So p is the variable, p is like x if f is constant.
00:38
So f is a constant here.
00:41
All right, so i'm going to rewrite this equation as f to the minus one equals p to the minus one plus q to the minus one.
00:48
And i'm taking the derivative with respect to p.
00:54
F is a constant, so the reciprocal of f is a constant.
00:58
So its derivative is zero.
01:01
P's derivative is minus one, p to the minus two, times the derivative of the of p with respect to p that would be one plus q is a function of p so its derivative is minus 1 q to the minus 2 times the derivative of q with respect to p okay so that dq dp is what we're trying to solve for so i'm going to get all these negative exponents out of here first all right i'm going to multiply everything by p squared q squared to get rid of the fractions so i get 0 equals minus q squared minus p squared dqdp...