Mathematics: Algebra, Geometry, and Calculus

Comparing Volume and Surface Area of Two Cylinders A and B

1. Setting up the comparison

Let cylinder A have radius $r_1$ and height $h_1$, and cylinder B have radius $r_2$ and height $h_2$.

$V = \pi r^2 h \qquad \text{(Volume)}$

$T.S.A. = 2\pi r(r + h) \qquad \text{(Total Surface Area)}$

$C.S.A. = 2\pi r h \qquad \text{(Curved Surface Area)}$

2. Method of comparison

• Compute $V_A$ and $V_B$, then compare by ratio $\dfrac{V_A}{V_B} = \dfrac{r_1^2 h_1}{r_2^2 h_2}$.
• Compute $T_A$ and $T_B$, then compare by ratio $\dfrac{T_A}{T_B} = \dfrac{r_1(r_1 + h_1)}{r_2(r_2 + h_2)}$.
• The ratio form is better than subtraction because it shows how many times one is larger, independent of units.

3. Impact of altering the size

Key conclusions:

1. Volume changes with the square of the radius but only the first power of the height — so radius has a much stronger influence on volume than height.
2. Curved surface area changes in direct proportion to both $r$ and $h$ (doubling either doubles C.S.A.).
3. Total surface area is not proportional to volume. When a cylinder is enlarged uniformly by factor $k$, volume grows as $k^3$ but surface area only as $k^2$. Hence bigger cylinders have more volume per unit of surface area — this is why large containers lose heat slowly and why cells cannot grow indefinitely large.

4. Numerical illustration

If $r_1 = 3$ cm, $h_1 = 5$ cm and $r_2 = 6$ cm, $h_2 = 5$ cm:

$\frac{V_A}{V_B} = \frac{3^2 \times 5}{6^2 \times 5} = \frac{9}{36} = \frac14$

$\frac{T_A}{T_B} = \frac{3(3+5)}{6(6+5)} = \frac{24}{66} = \frac{4}{11}$

So doubling the radius makes the volume 4 times but the total surface area only about 2.75 times — confirming that volume grows faster than surface area.

a) Hypothesis:
A hypothesis is a tentative statement or assumption made about a situation, which is to be tested or verified. In geometry, the condition after “if” in a conditional statement is called its hypothesis.

Properties of isometric transformation:

• It preserves distance between any two points.
• It preserves angles.
• It preserves shape and size, so the image is congruent to the original figure.
• It preserves area and perimeter.
• It preserves collinearity, betweenness, parallelism and perpendicularity.
• Examples: translation, rotation and reflection.

Properties of non-isometric transformation:

• It is not an isometry, so it does not preserve all distances.
• It changes size and/or shape of the figure.
• The image may be similar but not congruent; in dilation it is similar, while in shear or stretch it may not be similar.
• In dilation, angles are preserved but lengths change; in shear or stretch, angles may also change.
• Area may change; for dilation by scale factor $k$, area changes by factor $k^2$. A shear preserves area.
• Examples: dilation, shear and stretch.

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b) Solve by row equivalent method:

Given equations: $2x-y+4z=-3$ $x-4z=5$ $6x+y+2z=10$

Augmented matrix:

$$\left[\begin{array}{ccc|c} 2 & -1 & 4 & -3\\ 1 & 0 & -4 & 5\\ 6 & 1 & 2 & 10 \end{array}\right]$$

Apply row operations:

$R_1 \leftrightarrow R_2$:

$$\left[\begin{array}{ccc|c} 1 & 0 & -4 & 5\\ 2 & -1 & 4 & -3\\ 6 & 1 & 2 & 10 \end{array}\right]$$

$R_2 \to R_2-2R_1,\quad R_3 \to R_3-6R_1$:

$$\left[\begin{array}{ccc|c} 1 & 0 & -4 & 5\\ 0 & -1 & 12 & -13\\ 0 & 1 & 26 & -20 \end{array}\right]$$

$R_2 \to -R_2$:

$$\left[\begin{array}{ccc|c} 1 & 0 & -4 & 5\\ 0 & 1 & -12 & 13\\ 0 & 1 & 26 & -20 \end{array}\right]$$

$R_3 \to R_3-R_2$:

$$\left[\begin{array}{ccc|c} 1 & 0 & -4 & 5\\ 0 & 1 & -12 & 13\\ 0 & 0 & 38 & -33 \end{array}\right]$$

$R_3 \to \frac{R_3}{38}$:

$$\left[\begin{array}{ccc|c} 1 & 0 & -4 & 5\\ 0 & 1 & -12 & 13\\ 0 & 0 & 1 & -\frac{33}{38} \end{array}\right]$$

$R_2 \to R_2+12R_3$:

$$\left[\begin{array}{ccc|c} 1 & 0 & -4 & 5\\ 0 & 1 & 0 & \frac{49}{19}\\ 0 & 0 & 1 & -\frac{33}{38} \end{array}\right]$$

$R_1 \to R_1+4R_3$:

$$\left[\begin{array}{ccc|c} 1 & 0 & 0 & \frac{29}{19}\\ 0 & 1 & 0 & \frac{49}{19}\\ 0 & 0 & 1 & -\frac{33}{38} \end{array}\right]$$

Hence, $x=\frac{29}{19},\qquad y=\frac{49}{19},\qquad z=-\frac{33}{38}$

So the solution is: $\boxed{x=\frac{29}{19},\ y=\frac{49}{19},\ z=-\frac{33}{38}}$

a) Derivative from first principle

Let
[ f(x)=\frac{1}{\sqrt{2x+1}}=(2x+1)^{-1/2}. ]

By first principle, [ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}. ]

So, [ f'(x)=\lim_{h\to 0}\frac{\frac{1}{\sqrt{2x+2h+1}}-\frac{1}{\sqrt{2x+1}}}{h}. ]

Taking LCM, [ f'(x)=\lim_{h\to 0}\frac{\sqrt{2x+1}-\sqrt{2x+2h+1}}{h\sqrt{2x+2h+1}\sqrt{2x+1}}. ]

Rationalizing the numerator, [ f'(x)=\lim_{h\to 0}\frac{(2x+1)-(2x+2h+1)} {h\sqrt{2x+2h+1}\sqrt{2x+1}\left(\sqrt{2x+1}+\sqrt{2x+2h+1}\right)}. ]

[ =\lim_{h\to 0}\frac{-2h} {h\sqrt{2x+2h+1}\sqrt{2x+1}\left(\sqrt{2x+1}+\sqrt{2x+2h+1}\right)}. ]

Cancelling (h), [ f'(x)=\lim_{h\to 0}\frac{-2} {\sqrt{2x+2h+1}\sqrt{2x+1}\left(\sqrt{2x+1}+\sqrt{2x+2h+1}\right)}. ]

Putting (h=0), [ f'(x)=\frac{-2}{\sqrt{2x+1}\sqrt{2x+1}\left(\sqrt{2x+1}+\sqrt{2x+1}\right)} ] [ =\frac{-2}{(2x+1)(2\sqrt{2x+1})} =-\frac{1}{(2x+1)^{3/2}}. ]

[ \boxed{\frac{d}{dx}\left(\frac{1}{\sqrt{2x+1}}\right)=-\frac{1}{(2x+1)^{3/2}}} ]

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b) Equation of line of regression of (Y) on (X)

Let (n=7).

Thus, [ \Sigma X=109,\quad \Sigma Y=145,\quad \Sigma X^2=2189,\quad \Sigma XY=2984. ]

The regression coefficient of (Y) on (X) is [ b_{yx}=\frac{n\Sigma XY-\Sigma X\Sigma Y}{n\Sigma X^2-(\Sigma X)^2}. ]

Substituting values, [ b_{yx}=\frac{7(2984)-109(145)}{7(2189)-(109)^2} ] [ =\frac{20888-15805}{15323-11881} =\frac{5083}{3442}. ]

[ \boxed{b_{yx}=1.4768} ]

Now, [ \bar X=\frac{109}{7},\qquad \bar Y=\frac{145}{7}. ]

The intercept is [ a=\bar Y-b_{yx}\bar X ] [ a=\frac{145}{7}-\frac{5083}{3442}\cdot \frac{109}{7} ] [ a=-2.2809. ]

Hence the line of regression of (Y) on (X) is [ \boxed{Y=-2.2809+1.4768X} ]

or exactly, [ \boxed{Y=-\frac{7851}{3442}+\frac{5083}{3442}X} ]

or, [ \boxed{3442Y=5083X-7851} ]

Also, in deviation form: [ \boxed{Y-\frac{145}{7}=\frac{5083}{3442}\left(X-\frac{109}{7}\right)} ]