Day: October 13, 2020

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202010132314 Homework 1 (Q1)

The radius r of a right circular cylinder is decreasing at a rate of 12\,\mathrm{cm\, s^{-1}}, while its height h is decreasing at a rate of 25\,\mathrm{cm\, min^{-1}}. How is the volume changing when r=180\,\mathrm{cm} and h=500\,\mathrm{cm}? Is the volume increasing or decreasing?

Solution.

The volume of a right circular cylinder is calculated by the formula

V=\pi r^2h.

The volume V (a dependent variable) of a cylinder varies with its radius r and height h (both independent variables). The change of volume, simply put it, is a derivative of volume V with respect to time t:

\displaystyle{\frac{\mathrm{d}V}{\mathrm{d}t}}.

Differentiate V(r,h,t) wrt. time t:

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(r,h,t) & = \frac{\mathrm{d}}{\mathrm{d}t} (\pi r^2 h) \\ & = \pi \frac{\mathrm{d}}{\mathrm{d}t} (r^2h) \\ & = \pi \bigg[ r^2 \frac{\mathrm{d}}{\mathrm{d}t}(h) + h \frac{\mathrm{d}}{\mathrm{d}t} (r^2) \bigg] \\ & = \pi \bigg[ \big( r(t) \big)^2 \frac{\mathrm{d}}{\mathrm{d}t}\big( h(t) \big) + \big( h(t)\big) \bigg( \frac{\mathrm{d}}{\mathrm{d}t}\big( r(t) \big)^2 \bigg) \bigg] \\ & = \pi (r^2h' +2hrr' ) \\ & = \pi r^2h' + 2\pi hrr' \\ \end{aligned}

Given r=180\,\mathrm{cm}, h=500\,\mathrm{cm}, r'=-12\,\mathrm{cm\, s^{-1}}=-720\,\mathrm{cm\, min^{-1}}, and h'=-25\,\mathrm{cm\, min^{-1}}, you would have it.

In some formalism of partial derivatives,

\begin{aligned} \frac{\mathrm{d}V(r,h)}{\mathrm{d}t} & = \frac{\partial V}{\partial r}\frac{\mathrm{d}r}{\mathrm{d}t} + \frac{\partial V}{\partial h}\frac{\mathrm{d}h}{\mathrm{d}t} \\ & = (2\pi hr)\frac{\mathrm{d}r}{\mathrm{d}t} + (\pi r^2)\frac{\mathrm{d}h}{\mathrm{d}t}\\ & = 2\pi hrr' +\pi r^2h' \\ \end{aligned}

you could have it also.

Afterthought.

It just so happens that there are two lines of attack, by taking total/ordinary derivatives and by taking partial derivatives. Is here anyhow the difference? Is there anything the matter?

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202010130604 Example 1, Chapter 1.1, Methods in Physics II (2015-2016 Lectures)

Find the trajectory of a cannon ball fired by a cannoneer at 45^\circ above his eye level with an initial speed v.

Ans.

Imagine you are in his place, like the picture below:

Then, suppose you are opening fire at such positive \textrm{(+ve)\enspace }x-axis direction as that follows along the meridian. In addition, assume that the earth is to be lying flat beneath the cannon ball during its flight, that every pole (z-axis) upheld would be so much right-angled (/normal /perpendicular) to the ground as parallel (-transported) anywhere.

We neglect air friction.

1. Along the x-direction, the speed is kept v\cos 45^\circ.

2. Along the z-direction, the speed is initially v\sin 45^\circ. By the laws of gravity, the cannon ball will experience a net force m\mathbf{g} due to gravitational pull by the Earth. If upward direction is taken the positive sign, an equation of motion due to the Galilean transformation (i.e., v=u+at) will depict that v_z(t)=v\sin 45^\circ -gt.

3. One another equation s(t)=ut+\displaystyle{\frac{1}{2}}at^2 depicts how distance s (i.e., the magnitude of displacement \mathbf{s}) varies with time t during linear motion in constant acceleration a.

One might have already noticed I am setting the original question aside, as it were. Let’s pinpoint the answer now.

The trajectory should be obtained in the form:

\mathbf{s}(t)=s_x(t)\,\hat{\mathbf{i}} + s_z(t)\,\hat{\mathbf{k}}.

where

\begin{aligned} s_x(t) & = u_x t+\frac{1}{2}a_xt^2 \\ & = (v\cos 45^\circ )\, t + \frac{1}{2}(0)(t^2) \\ & = (v\cos 45^\circ )\, t \\ \end{aligned}

and

\begin{aligned} s_z(t) & = u_zt + \frac{1}{2}a_zt^2 \\ & = (v\sin 45^\circ )\, t + \frac{1}{2}(-g)(t^2) \\ \end{aligned}