Q. 4 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1.Given Information

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

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

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

$$\begin{gathered} \frac{d}{dt}\left(r\right(t\left)\right)=\frac{d}{dt}(3{\cos }^{2}t\mathbf{i}+5t\mathbf{j}+\frac{t}{t^2+1}\mathbf{k}) \\ \frac{d}{dt}\left(r\right(t\left)\right)=3\left(\frac{d}{dt}{\cos }^{2}t\right)\mathbf{i}+5\left(\frac{d}{dt}t\right)\mathbf{j}+\left(\frac{d}{dt}\frac{t}{t^2+1}\right)\mathbf{k} \\ \frac{d}{dt}\left(r\right(t\left)\right)=3A\mathbf{i}+5B\mathbf{j}+C\mathbf{k} \end{gathered}$$

Step 3. Firstly find the value of A = d d t cos 2 t

$$\begin{gathered} A=\frac{d}{dt}{\cos }^{2}t \\ A=2\cos t\left(-\sin t\right) \\ A=-2\cos t\sin t \end{gathered}$$

Step 4. Now finding the value of B = d d t t

$$\begin{gathered} B=\frac{d}{dt}t \\ B=1 \end{gathered}$$

Step 5. Now finding the value of C = d d t t t 2 + 1

$$\begin{gathered} \mathrm{Applying} \mathrm{the} \mathrm{quotient} \mathrm{rule} \mathrm{first}, \mathrm{we} \mathrm{have} \\ C=\frac{d}{dt}\frac{t}{t^2+1} \\ C=\frac{\frac{d}{dt}t·(t^2+1)-t·\frac{d}{dt}(t^2+1)}{t^2+1} \\ C=\frac{1·(t^2+1)-t·2t}{t^2+1} \\ C=\frac{t^2+1-2t^2}{t^2+1} \\ C=\frac{1-t^2}{t^2+1} \end{gathered}$$

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

$$\begin{gathered} \frac{d}{dt}\left(r\right(t\left)\right)=3(-2\cos t\sin t)\mathbf{i}+5\times 1\mathbf{j}+\frac{1-t^2}{t^2+1}\mathbf{k} \\ \frac{d}{dt}\left(r\right(t\left)\right)=-6\cos t\sin t\mathbf{i}+5\mathbf{j}+\frac{1-t^2}{t^2+1}\mathbf{k} \end{gathered}$$