# Find parametric equations for the tangent line to the curve

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.
$$\displaystyle{x}={1}+{2}\sqrt{{{t}}},\ {y}={t}^{{3}}-{t},\ {z}={t}^{{3}}+{t};\ {\left({3},{0},{2}\right)}$$

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Heack1991

We're given parametric equation $$\displaystyle{x}={1}+{2}\sqrt{{{t}}},\ {y}={t}^{{3}}-{t}$$ and $$\displaystyle{z}={t}^{{3}}+{t}$$ and we're asked to solve for the parametric equations of the tangent line to the curve at the point (3,0,2)
Given parametric equations, we know that the vector equatiom tr(t) is equal to
$$\displaystyle{r}{\left({t}\right)}={<}{1}+{2}\sqrt{{{t}}},{t}^{{3}}-{t},{t}^{{3}}+{t}{>}$$
Solve for $$\displaystyle{r}'{\left({t}\right)}$$ by differentiating each of the components of r(t) with respect to t
$$\displaystyle{r}'{\left({t}\right)}={<}{\frac{{{1}}}{{\sqrt{{{t}}}}}},{3}{t}^{{2}}-{1},{3}{t}^{{2}}+{1}{>}$$
The parameter value corresponding to $$\displaystyle{\left({3},{0},{2}\right)}$$ is t=1. Plug in t=1 into $$\displaystyle{r}'{\left({t}\right)}$$ to solve for $$\displaystyle{r}'{\left({1}\right)}$$
$$\displaystyle{r}'{\left({1}\right)}={<}{\frac{{{1}}}{{\sqrt{{{1}}}}}},{3}{\left({1}\right)}^{{2}}-{1},{3}{\left({1}\right)}^{{2}}+{1}{>}$$
$$\displaystyle={<}{1},{2},{4}{>}$$
Recall from the textbook that parametric equations for a line through the point $$\displaystyle{\left({x}_{{0}},{y}_{{0}},{z}_{{0}}\right)}$$ and parallel to the direction vector $$\displaystyle{<}{a},{b},{c}{>}$$ are
$$\displaystyle{x}={x}_{{0}}+{a}{t}\ {y}={y}_{{0}}+{b}{t}\ {z}={z}_{{0}}+{c}{t}$$
Substitute $$\displaystyle{\left({x}_{{0}},{y}_{{0}},{z}_{{0}}\right)}={\left({3},{0},{2}\right)}$$ and $$\displaystyle{<}{a},{b},{c}\ge{<}{1},{2},{4}{>}$$ into x, y and z, respectively to solve for the parametric equations of the tangent line to the curve
$$\displaystyle{x}={3}+{\left({1}\right)}{t}={3}+{t}\ {y}={\left({0}\right)}+{\left({2}\right)}{t}={2}{t}\ {z}={\left({2}\right)}+{\left({4}\right)}{t}={2}+{4}{t}$$