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Q. 4

Expert-verified
Found in: Page 1095

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Calculus of vector-valued functions: Calculate each of the following.$\frac{d}{dt}\left(r\left(t\right)\right),\mathrm{where}r\left(t\right)=3{\mathrm{cos}}^{2}t\mathbf{i}+5t\mathbf{j}+\frac{t}{{t}^{2}+1}\mathbf{k}$

The value of $\frac{d}{dt}\left(r\left(t\right)\right)=-6\mathrm{cos}t\mathrm{sin}t\mathbf{i}+5\mathbf{j}+\frac{1-{t}^{2}}{{t}^{2}+1}\mathbf{k}$

See the step by step solution

## Step 1.Given Information

Calculus of vector-valued functions: Calculate each of the following.

$\frac{d}{dt}\left(r\left(t\right)\right),\mathrm{where}r\left(t\right)=3{\mathrm{cos}}^{2}t\mathbf{i}+5t\mathbf{j}+\frac{t}{{t}^{2}+1}\mathbf{k}$

## Step 2. Now finding the value of ddt(r(t)).

$\frac{d}{dt}\left(r\left(t\right)\right)=\frac{d}{dt}\left(3{\mathrm{cos}}^{2}t\mathbf{i}+5t\mathbf{j}+\frac{t}{{t}^{2}+1}\mathbf{k}\right)\phantom{\rule{0ex}{0ex}}\frac{d}{dt}\left(r\left(t\right)\right)=3\left(\frac{d}{dt}{\mathrm{cos}}^{2}t\right)\mathbf{i}+5\left(\frac{d}{dt}t\right)\mathbf{j}+\left(\frac{d}{dt}\frac{t}{{t}^{2}+1}\right)\mathbf{k}\phantom{\rule{0ex}{0ex}}\frac{d}{dt}\left(r\left(t\right)\right)=3A\mathbf{i}+5B\mathbf{j}+C\mathbf{k}$

## Step 3. Firstly find the value of A=ddtcos2t

$A=\frac{d}{dt}{\mathrm{cos}}^{2}t\phantom{\rule{0ex}{0ex}}A=2\mathrm{cos}t\left(-\mathrm{sin}t\right)\phantom{\rule{0ex}{0ex}}A=-2\mathrm{cos}t\mathrm{sin}t$

## Step 4. Now finding the value of B=ddtt

$B=\frac{d}{dt}t\phantom{\rule{0ex}{0ex}}B=1$

## Step 5. Now finding the value of C=ddttt2+1

$\mathrm{Applying}\mathrm{the}\mathrm{quotient}\mathrm{rule}\mathrm{first},\mathrm{we}\mathrm{have}\phantom{\rule{0ex}{0ex}}C=\frac{d}{dt}\frac{t}{{t}^{2}+1}\phantom{\rule{0ex}{0ex}}C=\frac{\frac{d}{dt}t·\left({t}^{2}+1\right)-t·\frac{d}{dt}\left({t}^{2}+1\right)}{{t}^{2}+1}\phantom{\rule{0ex}{0ex}}C=\frac{1·\left({t}^{2}+1\right)-t·2t}{{t}^{2}+1}\phantom{\rule{0ex}{0ex}}C=\frac{{t}^{2}+1-2{t}^{2}}{{t}^{2}+1}\phantom{\rule{0ex}{0ex}}C=\frac{1-{t}^{2}}{{t}^{2}+1}$

## Step 6 Now putting the value of A,B,C.

$\frac{d}{dt}\left(r\left(t\right)\right)=3\left(-2\mathrm{cos}t\mathrm{sin}t\right)\mathbf{i}+5×1\mathbf{j}+\frac{1-{t}^{2}}{{t}^{2}+1}\mathbf{k}\phantom{\rule{0ex}{0ex}}\frac{d}{dt}\left(r\left(t\right)\right)=-6\mathrm{cos}t\mathrm{sin}t\mathbf{i}+5\mathbf{j}+\frac{1-{t}^{2}}{{t}^{2}+1}\mathbf{k}$