Find the second derivative of the vector-valued function →r(t)=(5t+7sin⁡(t))i+(7t+2cos⁡(t))j

Step 1 Given that: r(t)=(5t+7sin⁡t)i+(7t+2cos⁡t)j So, 1st derivative of r: drdt=iddt(5t+7×sin⁡t)+j×ddt(7t+2cos⁡t) =i(5ddt(t)+7ddt(sin⁡t))+j(7ddt(t)+2ddtcos⁡t) =i(5+7cos⁡t)+j(7−2sin⁡t) So, drdt=(5+7cos⁡t)i+(7−2sin⁡t)j 2nd derivative of r: d2rdt2=ddt[(5+7cos⁡t)i+(7−2sin⁡t)j] =(5ddt(1)+7ddtcos⁡t)i+(7ddt(1)−2ddtsin⁡t)j =(0−7sin⁡t)i+(0−2cos⁡t)j −7sin⁡t×i−2cos⁡t×j So, d2rdt2=−7sin⁡t×i−2cos⁡t×j

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deliredejoker7m

Answered question

2021-11-28

Find the second derivative of the vector-valued function

\to r (t) = (5 t + 7 \sin (t)) i + (7 t + 2 \cos (t)) j

Answer & Explanation

Befoodly

Beginner2021-11-29Added 17 answers

Step 1
Given that:

r (t) = (5 t + 7 \sin t) i + (7 t + 2 \cos t) j

So, 1st derivative of r:

\frac{d r}{d t} = i \frac{d}{d t} (5 t + 7 \times \sin t) + j \times \frac{d}{d t} (7 t + 2 \cos t)

= i (5 \frac{d}{d t} (t) + 7 \frac{d}{d t} (\sin t)) + j (7 \frac{d}{d t} (t) + 2 \frac{d}{d t} \cos t)

= i (5 + 7 \cos t) + j (7 - 2 \sin t)

So,

\frac{d r}{d t} = (5 + 7 \cos t) i + (7 - 2 \sin t) j

2nd derivative of r:

\frac{d^{2} r}{{d t}^{2}} = \frac{d}{d t} [(5 + 7 \cos t) i + (7 - 2 \sin t) j]

= (5 \frac{d}{d t} (1) + 7 \frac{d}{d t} \cos t) i + (7 \frac{d}{d t} (1) - 2 \frac{d}{d t} \sin t) j

= (0 - 7 \sin t) i + (0 - 2 \cos t) j

- 7 \sin t \times i - 2 \cos t \times j

So,

\frac{d^{2} r}{{d t}^{2}} = - 7 \sin t \times i - 2 \cos t \times j

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