Step by step solution to "Derivatives of vector-valued functions Differentiate the following function...."

osteoblogda

Answered question

2021-12-06

Derivatives of vector-valued functions Differentiate the following function.

r (t) = < 4, 3 \cos 2 t, 2 \sin 3 t >

soanooooo40

Beginner2021-12-07Added 35 answers

Step 1
the given function is:

r (t) = < 4, 3 \cos (2 t), 2 \sin (3 t) >

we have to differentiate the given function.
the given function can be written as:

r (t) = 4 i + 3 \cos (2 t) j + 2 \sin (3 t) k

Step 2
therefore we have

r (t) = 4 i + 3 \cos (2 t) j + 2 \sin (3 t) k

now differentiate both the sides with respect to t.
therefore,

\frac{d (r (t))}{d t} = \frac{d (4 i + 3 \cos (2 t) j + 2 \sin (3 t) k)}{d t}

= \frac{d 4}{d t} i + \frac{d 3 \cos (2 t)}{d t} j + \frac{d 2 \sin (3 t)}{d t} k

= 0 i + 3 \frac{d \cos (2 t)}{d t} j + 2 \frac{d \sin (3 t)}{d t} k

= 3 (- \sin (2 t)) \frac{d 2 t}{d t} j + 2 (\cos (3 t)) \frac{d 3 t}{d t} k

= - 3 \sin (2 t) \times 2 j + 2 \cos (3 t) \times 3 k

= - 6 \sin (2 t) j + 6 \cos (3 t) k

Step 3
therefore ,

\frac{d (r (t))}{d t} = - 6 \sin (2 t) j + 6 \cos (3 t) k

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?