"Determine the equation of the tangent line to..." solved and explained

shadsiei

Answered question

2021-10-28

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

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

Tasneem Almond

Skilled2021-10-29Added 91 answers

Remember that for a path c(t) the equation of its tangent line at the point

c (t_{0})

is given with:

l (t) = c (t_{0}) + c^{'} (t_{0}) (t - t_{0})

In our case

c (t) = (\sin 3 t, \cos 3 t, 2 t^{\frac{5}{2}})

and

t_{0} = 1

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

c^{'} (t) = ({(\sin 3 t)}^{'}, {(\cos 3 t)}^{'}, {(2 t^{\frac{5}{2}})}^{'}) = (3 \cos 3 t, - 3 \sin 3 t, 5 t^{\frac{3}{2}})

Which gives:

c^{'} (1) = (3 \cos 3, - 3 \sin 3, 5)

Finally, using (1) we have:

l (t) = c (1) + c^{'} (1) (t - 1)

I.e.

l (t) = (\sin 3, \cos 3, 2) + (3 \cos 3 - 3 \sin 3, 5) (t - 1)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?