Question

2. For $w(t)$ taking values in $C$, define a new curve by $z(s) = w(t)^2$, $\frac{ds}{dt} = |w(t)|^2$. Show that if $w''(t) = -w(t)$, then $z''(s) = -4E \frac{z(s)}{|z(s)|^3}$, so $z(s)$ solves the Kepler problem.

          2. For $w(t)$ taking values in $C$, define a new curve by
$z(s) = w(t)^2$, $\frac{ds}{dt} = |w(t)|^2$.
Show that if $w''(t) = -w(t)$, then
$z''(s) = -4E \frac{z(s)}{|z(s)|^3}$,
so $z(s)$ solves the Kepler problem.

2. For w(t) taking values in C, define a new curve by
z(s) = w(t)^2, (ds)/(dt) = |w(t)|^2.
Show that if w”(t) = -w(t), then
z”(s) = -4E (z(s))/(|z(s)|^3),
so z(s) solves the Kepler problem.

Added by Michaela R.

Calculus

Ron Larson, Robert P. Hostetler, Bruce… 8th Edition

Chapter 11

Instant Answer

Solved by Expert Ekaveera Kumar

Step 1

This means that w and t have opposite signs. If we substitute -wt for wt in the equations above, we get: s = (-wt)^2 - 91 s = w^2t^2 - 91 s = (-wt)^12 s = w^12t^12 Now, let's simplify these equations: s = w^2t^2 - 91 s = w^12t^12 Show more…

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