October 17, 2023 – Physicspupil

Evaluate

$\displaystyle{\int_{0}^2}\frac{x^2\,\mathrm{d}x}{\sqrt{9-x^3}}$ .

Roughwork.

Let

$\begin{aligned} f(x) & =x^2 \\ g(x) & = \sqrt{9-x^3} \\ \end{aligned}$

s.t.

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}x} & = 2x \\ \frac{\mathrm{d}g}{\mathrm{d}x} & = -\frac{3}{2}\bigg(\frac{x^2}{\sqrt{9-x^3}}\bigg) \\ & = -\frac{3}{2}\frac{f}{g} \\ -\frac{2}{3}\,\mathrm{d}g & = \frac{f}{g}\,\mathrm{d}x \\ \end{aligned}$

Then

$\begin{aligned} &\quad\enspace \int_{0}^{2}\frac{x^2\,\mathrm{d}x}{\sqrt{9-x^3}} \\ & = \int_{0}^{2}\frac{f(x)}{g(x)}\,\mathrm{d}x \\ & = -\frac{2}{3}\int_{0}^{2}\mathrm{d}g \\ & = -\frac{2}{3} \big[ g(x)\Big]\Big|_{0}^{2} \\ & = -\frac{2}{3} \Big[\sqrt{9-x^3}\Big]\Big|_{0}^{2} \\ & = -\frac{2}{3}(1-3) \\ & = \frac{4}{3} \\ \end{aligned}$

Ah, but no sense of physical meaning.

This problem is not to be attempted.

A particle moves along a straight line so that its distance ( $s$ ) from a fixed point $O$ on the line after $t\,\mathrm{s}$ is given by

$s=t^3-12t^2+45t$ .

Find the distances from $O$ when the particle is momentarily at rest and the accelerations at these times.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI units}

Roughwork.

We give the direction to the right a positive sign.

By substituting $t=0$ for time, we found the original position of the particle to be at a distance:

$\begin{aligned} s(t=0) & = |\mathbf{s}(0)| \\ & = |(0)^3-12(0)^2+45(0)\,\hat{\mathbf{i}}| \\ & = 0\,\mathrm{unit} \\ \end{aligned}$

from point $O$ , which shall be called the origin hence.

The instantaneous velocity being

$\begin{aligned} \mathbf{v}(t)& =\lim_{\Delta t\to 0}\frac{\Delta \mathbf{s}}{\Delta t} \\ & = \frac{\mathrm{d}\mathbf{s}}{\mathrm{d}t} \\ & = \frac{\mathrm{d}}{\mathrm{d}t}(t^3-12t^2+45t)\,\hat{\mathbf{i}} \\ & = (3)t^{(3)-1} - 12(2)t^{(2)-1} + 45(1)t^{(1)-1} \\ & = 3t^2 -24t + 45\,\hat{\mathbf{i}} \\ \end{aligned}$

the average velocity during time interval of $t\in [0,t']$ being

$\begin{aligned} v_{\textrm{avg}}(t') & = \frac{s(t')}{t'} \\ & = \frac{(t')^3-12(t')^2+45(t')}{t'} \\ & = t'^2-12t'+45 \\ \end{aligned}$

in magnitude; the instantaneous acceleration being

$\begin{aligned} a(t) & = \bigg| \frac{\mathrm{d}^2}{\mathrm{d}t^2}(s(t))\,\hat{\mathbf{i}}\bigg| \\ & = \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}}{\mathrm{d}t}(t^3-12t^2+45t) \bigg) \\ & = \frac{\mathrm{d}}{\mathrm{d}t}(3t^2-24t+45) \\ & = 6t-24\\ \end{aligned}$