Answer to Question #241456 in Calculus for Akki

Question

Show that the two curves (e', e², 1-e) and (1-theta , cos theta , sin theta ) intersect at the point (1, 1, 0). What is the angle between their tangents at that point?

Answer 1

For the two curves to intersect at (1,1,0), then the value of each poianat of the curve must be the same and equal to (1,1,0)

Suppose $(e^t,e^{2t}, 1-e^t)=(1-\theta, \cos \theta, \sin \theta)\\ \implies t=0, \theta=0\\ \text{At } t=0\\ (e^t,e^{2t}, 1-e^t)=(1,1,0)\\ \text{At } \theta=0\\ (1-\theta, \cos \theta, \sin \theta)=(1,1,0)\\ \text{Hence, the two curves intersect at } (1,1,0).\\ \text{Let } r(t)=(e^t,e^{2t}, 1-e^t), ~~r(\theta)=(1-\theta, \cos \theta, \sin \theta)\\ r'(t)= (e^t, 2e^{2t}, -e^t)\\ r'(0)=(1,2,-1)\\ r'(\theta)=(-1, -\sin \theta, \cos \theta)\\ r'(0)=(-1,0,1)\\ \text{The angle between their tangent is }\\ \cos \alpha=\frac{(1,2,-1)\cdot(-1,0,1)}{(\sqrt{1^2+2^2+(-1)^2})\cdot \sqrt{(-1)^2+0^2+1^2}}\\ \cos \alpha=\frac{-2}{2\sqrt{3}}\\ \alpha=cos^{-1}(-\frac{1}{\sqrt{3}})\\ \alpha= 125.26°$