Determine the equation of the tangent line to the given path at the specified value of t.

$(\mathrm{sin}3t,\mathrm{cos}3t,2{t}^{\frac{5}{2}});t=1$

shadsiei
2021-10-28
Answered

Determine the equation of the tangent line to the given path at the specified value of t.

$(\mathrm{sin}3t,\mathrm{cos}3t,2{t}^{\frac{5}{2}});t=1$

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Tasneem Almond

Answered 2021-10-29
Author has **91** answers

Remember that for a path c(t) the equation of its tangent line at the point $c\left({t}_{0}\right)$ is given with:

$l\left(t\right)=c\left({t}_{0}\right)+{c}^{\prime}\left({t}_{0}\right)(t-{t}_{0})$

In our case$c\left(t\right)=(\mathrm{sin}3t,\mathrm{cos}3t,2{t}^{\frac{5}{2}})$ and ${t}_{0}=1$

To obtain c'(t) calculate the derivatives component-wise:

${c}^{\prime}\left(t\right)=({\left(\mathrm{sin}3t\right)}^{\prime},{\left(\mathrm{cos}3t\right)}^{\prime},{\left(2{t}^{\frac{5}{2}}\right)}^{\prime})=(3\mathrm{cos}3t,-3\mathrm{sin}3t,5{t}^{\frac{3}{2}})$

Which gives:

${c}^{\prime}\left(1\right)=(3\mathrm{cos}3,-3\mathrm{sin}3,5)$

Finally, using (1) we have:

$l\left(t\right)=c\left(1\right)+{c}^{\prime}\left(1\right)(t-1)$

I.e.

$l\left(t\right)=(\mathrm{sin}3,\mathrm{cos}3,2)+(3\mathrm{cos}3-3\mathrm{sin}3,5)(t-1)$

In our case

To obtain c'(t) calculate the derivatives component-wise:

Which gives:

Finally, using (1) we have:

I.e.

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