Answer to Question #233707 in Differential Geometry | Topology for JOHNNIE

Question

Let the position vector of a stone at time,t be given as r(t)=cosh (t^2-1)I+ sinh(1-t)j+Bt^2k

Assume that the position vector is Normal to the acceleration vector.

Find the value of B at Time t=5seconds

Answer 1

$$r(t)=\cosh (t^2-1)i+ \sinh(1-t)j+Bt^2k$$

$$v(t)=r'(t)$$

$$=2t\sinh (t^2-1)i- \cosh(1-t)j+2Btk$$

$$a(t)=r''(t)$$

$$=(2\sinh (t^2-1)+4t^2\cosh (t^2-1))i+ \sinh(1-t)j+2Bk$$

$$r(t)\cdot a(t)=0$$

$$\cosh (t^2-1)(2\sinh (t^2-1)+4t^2\cosh (t^2-1))$$

$$+ \sinh(1-t) \sinh(1-t)+Bt^2(2B)=0$$

$$B^2t^2=-\cosh (t^2-1)\sinh (t^2-1)$$

$$-2t^2\cosh^2 (t^2-1)-\sinh^2(1-t)$$

$t=5$

$$25B^2=-\cosh (25-1)\sinh (25-1)$$

$$-2(25)\cosh^2 (25-1)-\sinh^2(1-5)$$

$$B^2=\dfrac{1}{25}(-\dfrac{\sinh(48)}{2}-50\cosh^2 (24)+\sinh^2(4))<0$$

There are no solution for $B\in \R.$