Solution;
1.Find r' and T for r=ht2 ,1,ti
r'= d d t \frac {d}{dt} d t d r=2ht,0,i
T=r ′ ∣ r ′ ∣ \frac{r'}{|r'|} ∣ r ′ ∣ r ′
∣ r ′ ∣ = ( 2 h t ) 2 + 0 2 + i 2 = 4 h 2 t 2 − 1 |r'|=\sqrt{(2ht)^2+0^2+i^2}=\sqrt{4h^2t^2-1} ∣ r ′ ∣ = ( 2 h t ) 2 + 0 2 + i 2 = 4 h 2 t 2 − 1
T= 2 h t 4 h 2 t 2 − 1 , 0 , i 4 h 2 t 2 − 1 \frac{2ht}{\sqrt{4h^2t^2-1}},0,\frac{i}{\sqrt{4h^2t^2-1}} 4 h 2 t 2 − 1 2 h t , 0 , 4 h 2 t 2 − 1 i
2.Find r' and T for r=hcost ,sin2t ,t2 i
r'= d d t \frac{d}{dt} d t d r=-hsint,2cos2t,2it
T=r ′ ∣ r ′ ∣ \frac{r'}{|r'|} ∣ r ′ ∣ r ′
∣ r ′ ∣ = ( − h s i n t ) 2 + ( 2 c o s 2 t ) 2 + ( 2 i t ) 2 |r'|=\sqrt{(-hsint)^2+(2cos2t)^2+(2it)^2} ∣ r ′ ∣ = ( − h s in t ) 2 + ( 2 cos 2 t ) 2 + ( 2 i t ) 2 =h 2 s i n 2 t + 4 c o s 2 ( 2 t ) − 4 t 2 \sqrt{h^2sin^2t+4cos^2(2t)-4t^2} h 2 s i n 2 t + 4 co s 2 ( 2 t ) − 4 t 2
T= − h s i n t h 2 s i n 2 t + 4 c o s 2 ( 2 t ) − 4 t 2 , \frac{-hsint}{\sqrt{h^2sin^2t+4cos^2(2t)-4t^2}}, h 2 s i n 2 t + 4 co s 2 ( 2 t ) − 4 t 2 − h s in t , 2 c o s 2 t h 2 s i n 2 t + 4 c o s 2 ( 2 t ) − 4 t 2 , \frac{2cos2t}{\sqrt{h^2sin^2t+4cos^2(2t)-4t^2}}, h 2 s i n 2 t + 4 co s 2 ( 2 t ) − 4 t 2 2 cos 2 t , 2 i t h 2 s i n 2 t + 4 c o s 2 ( 2 t ) − 4 t 2 \frac{2it}{\sqrt{h^2sin^2t+4cos^2(2t)-4t^2}} h 2 s i n 2 t + 4 co s 2 ( 2 t ) − 4 t 2 2 i t
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