Find parametric equations for the tangent line to the curve

podnescijy 2021-11-20 Answered
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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Expert Answer

Heack1991
Answered 2021-11-21 Author has 1095 answers

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
$$\displaystyle{x}={3}+{\left({1}\right)}{t}={3}+{t}\ {y}={\left({0}\right)}+{\left({2}\right)}{t}={2}{t}\ {z}={\left({2}\right)}+{\left({4}\right)}{t}={2}+{4}{t}$$

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