Question

If we have a vector as $r={{t}^{2}}i-tj+\left( 2t+1 \right)k$, find $\left| \dfrac{dr}{dt} \right|$.

Answer 1

In this problem, we have to find the differentiation of the given expression and then find the modulus of it. We can first differentiate the given expression using the differentiation formulas. Where we can use the formulas for each of the terms given separately. We can then find the modulus for the differentiated result to find the required answer.

Complete step by step solution:
Here we have to find the differentiation of the given expression and then find the modulus of it.
We know that the expression given to be differentiate is,
$\Rightarrow r={{t}^{2}}\overset{\hat{\ }}{\mathop{i}}\,-t\overset{\hat{\ }}{\mathop{j}}\,+\left( 2t+1 \right)\overset{\hat{\ }}{\mathop{k}}\,$
We can now differentiate the given expression with respect to t, we get
$\Rightarrow \dfrac{dr}{dt}=2t\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{j}}\,+2\overset{\hat{\ }}{\mathop{k}}\,$…….. (1)
Since, the first term is differentiated as ${{t}^{2}}dt=2t$
The second term is$tdt=1$ and the third term is $\left( 2t+1 \right)dt=2+0=2$.
We can now find the modulus for the expression (1), which will be the square root of the sum of the squared terms.
We know that the modulus of the general equation $a\overset{\hat{\ }}{\mathop{i}}\,+b\overset{\hat{\ }}{\mathop{j}}\,+c\overset{\hat{\ }}{\mathop{k}}\,$ is
$\Rightarrow \left| x \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$
We can now apply this in (1), we get
$\Rightarrow \left| \dfrac{dr}{dt} \right|=\sqrt{{{\left( 2t \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}}}$
We can now simplify the above step, we get
$\Rightarrow \left| \dfrac{dr}{dt} \right|=\sqrt{4{{t}^{2}}+5}$
Therefore, $\left| \dfrac{dr}{dt} \right|=\sqrt{4{{t}^{2}}+5}$.

Note: We should always remember the differentiating formula to solve these types of problems. We should also know that the modulus of the general equation $a\overset{\hat{\ }}{\mathop{i}}\,+b\overset{\hat{\ }}{\mathop{j}}\,+c\overset{\hat{\ }}{\mathop{k}}\,$ is $\left| x \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$ . We should also remember to simplify the terms inside the square root to find the answer for the given expression. We should know some of the differentiating formulas like ${{t}^{2}}dt=2t$ and $tdt=1$.