Find parametric equations of the straight line tangent to the following space curve at the point $P(−3,−9,0)$ on the curve. $$r(t)=(8t^2+63t+46)i+(2t^3-98t-9)j+\frac{70\sqrt3}{\pi}(1+2\cos(\frac{4\pi t}{21}))k$$
I found the tangent equation as $$x(t)=-71+15t$$ $$y(t)=-585+388t$$ $$z(t)=\frac{210\sqrt3}{\pi}$$ But I have to paramterize in a way such as $$x(t)=~-3- 49s$$ $$y(t)=-9+196s$$ $$z(t)= ~-40s$$ How should I do it?
$$\vec r(t)=\left(8 t^2+63 t+46,2 t^3-98 t-9,\frac{70 \sqrt{3} \left(2 \cos \left(\frac{4 \pi t}{21}\right)+1\right)}{\pi }\right)$$ gives the point $P(−3,−9,0)$ for $t=-7$
we have $$\vec r\,'(t)=\left(16 t+63,6 t^2-98,-\frac{80 \sin \left(\frac{4 \pi t}{21}\right)}{\sqrt{3}}\right)$$ plugging $t=-7$ we get $\vec t=(-49,\;196,\;-40)$
the tangent line has parametric equations $(x,y,z)=(−3,−9,0)+t(-49,\;196,\;-40)$
that is
$ \left\{ \begin{array}{l} x=-3-49 t \\ y=-9+196 t \\ z=-40 t \\ \end{array} \right. $
Hope this helps
