# Determine the equation of the tangent line to the given path at the specified va

Determine the equation of the tangent line to the given path at the specified value of t.
$\left(\mathrm{sin}3t,\mathrm{cos}3t,2{t}^{\frac{5}{2}}\right);t=1$
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Tasneem Almond
Remember that for a path c(t) the equation of its tangent line at the point $c\left({t}_{0}\right)$ is given with:
$l\left(t\right)=c\left({t}_{0}\right)+{c}^{\prime }\left({t}_{0}\right)\left(t-{t}_{0}\right)$
In our case $c\left(t\right)=\left(\mathrm{sin}3t,\mathrm{cos}3t,2{t}^{\frac{5}{2}}\right)$ and ${t}_{0}=1$
To obtain c'(t) calculate the derivatives component-wise:
${c}^{\prime }\left(t\right)=\left({\left(\mathrm{sin}3t\right)}^{\prime },{\left(\mathrm{cos}3t\right)}^{\prime },{\left(2{t}^{\frac{5}{2}}\right)}^{\prime }\right)=\left(3\mathrm{cos}3t,-3\mathrm{sin}3t,5{t}^{\frac{3}{2}}\right)$
Which gives:
${c}^{\prime }\left(1\right)=\left(3\mathrm{cos}3,-3\mathrm{sin}3,5\right)$
Finally, using (1) we have:
$l\left(t\right)=c\left(1\right)+{c}^{\prime }\left(1\right)\left(t-1\right)$
I.e.
$l\left(t\right)=\left(\mathrm{sin}3,\mathrm{cos}3,2\right)+\left(3\mathrm{cos}3-3\mathrm{sin}3,5\right)\left(t-1\right)$