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$4=2u\frac{du}{dt}$andlocalid="1649780328523" $u=2+3t,$find${\overline{)\frac{du}{dt}}}_{t=4}$

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

Step-by-step solution

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 ii

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 iii.

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 $t=4$

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