Post by Gary on Jun 17, 2017 14:03:22 GMT
Vasco,
In exercise 5, ch 8, at the top of p. 2, you have
$\hspace{5em}\int_K\frac{1}{z}dz = \int^b_a\frac{v(t)}{z(t)}dt$
Can you explain why this substitution works, particularly why v(t) works in the numerator?
Oh, nevermind, I see it now: $dz = v(t) dt$.
Gary
In exercise 5, ch 8, at the top of p. 2, you have
$\hspace{5em}\int_K\frac{1}{z}dz = \int^b_a\frac{v(t)}{z(t)}dt$
Can you explain why this substitution works, particularly why v(t) works in the numerator?
Oh, nevermind, I see it now: $dz = v(t) dt$.
Gary