Question

Find angle between the tangents at t = 1 and t = 2 to the curve 𝑟̅ = 𝑡2 𝑖̅+ (𝑡3 − 2𝑡) 𝑗̅ + (3𝑡 − 4) 𝑘̅.

Answer 1

$r'(t)=2ti+(3t^2-2)j+3k$

r'(1)=2i+ 1j+3k

r'(2)=4i+10j+3k

cos(r'(1),r'(2))=(2*4+1*10+3*3)/$(\sqrt{r'(1)}*\sqrt{r'(2)})$ =27/$(\sqrt{14}*\sqrt{125})$ =27/$(5\sqrt{70})$ angle=arcos(27/$(5\sqrt{70}))$ )