Q 2. TB - Calculus

Question

Evaluation in Leibniz notation: Given that $r=r\left(t\right), s=s\left(t\right)$ and $u=u\left(t\right)$ are functions of $t$, answer each of the following.

$\left(i\right)$ If $\frac{ds}{dt}=3s^2-4,$ find ${\overline{)\frac{ds}{dt}}}_{s=2}$

$\left(ii\right)$ If $r^2\frac{dr}{dt}-2r=0,$find ${\overline{)\frac{dr}{dt}}}_{r=3}$

$\left(iii\right)$ If $4=2u\frac{du}{dt}$ and $u=2+3t,$ find ${\overline{)\frac{du}{dt}}}_{t=4}$

Step-by-Step Solution

Answer

Part $\left(i\right) {\overline{)\frac{ds}{dt}}}_{s=2}=8$

  

Part $\left(ii\right) {\overline{)\frac{dr}{dt}}}_{r=3}=\frac{2}{3}$

          

Part $\left(iii\right) {\overline{)\frac{du}{dt}}}_{t=4}=\frac{1}{7}$

Part (i) Step 1. Definition.

Leibniz notation is a method for representing the derivative that uses the symbols $dx$ and $dy$ to designate infinitely small increments of $x$ and $y$.  

Part (i) Step 2. Explanation of part i

Given that $\frac{ds}{dt}=3s^2-4$

Put $s=2$

$\Rightarrow {\overline{)\frac{ds}{dt}}}_{s=2}=3{\left(2\right)}^2-4=8$

Part (ii) Step 1. Explanation of part i i

Given that $r^2\frac{dr}{dt}-2r=0$

$\Rightarrow \frac{dr}{dt}=\frac{2}{r}$

Put $r=3$

$\Rightarrow {\overline{)\frac{dr}{dt}}}_{r=3}=\frac{2}{3}$

Part (iii) Step 4. Explanation of part i i i .

Given that $4=2u\frac{du}{dt}$ and $u=2+3t$

$\Rightarrow 4=2\left(2+3t\right)\frac{du}{dt}$

$\Rightarrow \frac{du}{dt}=\frac{2}{2+3t}$

Put $t=4$

$\Rightarrow {\overline{)\frac{du}{dt}}}_{t=4}=\frac{2}{14}=\frac{1}{7}$