Vardaan Learning Institute

A particle moves along the curve $y = x^2 + 2x$. At what point on the curve are the $x$ and $y$ coordinates of the particle changing at the same rate? Solution: Given $\frac{dy}{dt} = \frac{dx}{dt}$. Differentiating the curve: $\frac{dy}{dt} = (2x + 2)\frac{dx}{dt}$.
Substitute $\frac{dy}{dt} = \frac{dx}{dt}$: $\frac{dx}{dt} = (2x + 2)\frac{dx}{dt} \implies 1 = 2x + 2 \implies 2x = -1 \implies x = -1/2$.
When $x = -1/2$, $y = (-1/2)^2 + 2(-1/2) = 1/4 - 1 = -3/4$.
The required point is $(-1/2, -3/4)$.

A $5\text{ m}$ long ladder is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2\text{ m/s}$. How fast is the angle $\theta$ between the ladder and the ground changing when the foot of the ladder is $4\text{ m}$ away from the wall? Solution: Let $x$ be the distance from the wall. We know $\cos\theta = \frac{x}{5}$.
Differentiating w.r.t $t$: $-\sin\theta \cdot \frac{d\theta}{dt} = \frac{1}{5}\frac{dx}{dt}$.
Given $\frac{dx}{dt} = 2\text{ m/s}$, and $x = 4\text{ m}$. By Pythagoras, height $y = \sqrt{25-16} = 3\text{ m}$. So, $\sin\theta = \frac{3}{5}$.
$- \left(\frac{3}{5}\right) \frac{d\theta}{dt} = \frac{1}{5}(2) \implies -3 \frac{d\theta}{dt} = 2 \implies \frac{d\theta}{dt} = -\frac{2}{3}\text{ rad/s}$.
The angle is decreasing at $\frac{2}{3}\text{ rad/s}$.

Water is leaking from a conical funnel at the rate of $5\text{ cm}^3\text{/sec}$. If the radius of the base of the funnel is $5\text{ cm}$ and its height is $10\text{ cm}$, find the rate at which the water level is dropping when the water is $2.5\text{ cm}$ deep. Solution: By similar triangles, $\frac{r}{h} = \frac{5}{10} \implies r = \frac{h}{2}$.
Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi}{12}h^3$.
Differentiating w.r.t $t$: $\frac{dV}{dt} = \frac{\pi}{4}h^2 \frac{dh}{dt}$. Given $\frac{dV}{dt} = -5$ (leaking) and $h = 2.5 = 5/2$.
$-5 = \frac{\pi}{4}\left(\frac{25}{4}\right)\frac{dh}{dt} \implies -5 = \frac{25\pi}{16}\frac{dh}{dt} \implies \frac{dh}{dt} = \frac{-80}{25\pi} = -\frac{16}{5\pi}\text{ cm/s}$.
The water level is dropping at $\frac{16}{5\pi}\text{ cm/s}$.

A man $2\text{ m}$ high walks at a uniform speed of $5\text{ km/h}$ away from a lamp-post $6\text{ m}$ high. Find the rate at which the length of his shadow increases. Solution: Let distance of man from pole be $x$ and length of shadow be $y$. Given $\frac{dx}{dt} = 5\text{ km/h}$.
By similar triangles: $\frac{6}{x+y} = \frac{2}{y} \implies 6y = 2x + 2y \implies 4y = 2x \implies 2y = x$.
Differentiating w.r.t $t$: $2\frac{dy}{dt} = \frac{dx}{dt} \implies 2\frac{dy}{dt} = 5 \implies \frac{dy}{dt} = 2.5\text{ km/h}$.
The shadow length increases at $2.5\text{ km/h}$.

Find the intervals in which the function $f(x) = \frac{x^4}{4} - x^3 - 5x^2 + 24x + 12$ is strictly increasing or strictly decreasing. Solution: $f'(x) = x^3 - 3x^2 - 10x + 24$. By trial, $x=2$ is a root ($8 - 12 - 20 + 24 = 0$).
Factoring gives $f'(x) = (x-2)(x^2 - x - 12) = (x-2)(x-4)(x+3)$.
Critical points are $-3, 2, 4$. Using the wavy curve method:
Intervals of Strict Increase ($f'(x) > 0$): $(-3, 2) \cup (4, \infty)$.
Intervals of Strict Decrease ($f'(x) < 0$): $(-\infty, -3) \cup (2, 4)$.

Find the intervals in which the function $f(x) = (x+1)^3(x-3)^3$ is strictly increasing. Solution: $f'(x) = 3(x+1)^2(x-3)^3 + 3(x-3)^2(x+1)^3 = 3(x+1)^2(x-3)^2[x-3+x+1] = 3(x+1)^2(x-3)^2(2x-2)$.
$f'(x) = 6(x-1)(x+1)^2(x-3)^2$.
Since $(x+1)^2 \ge 0$ and $(x-3)^2 \ge 0$, the sign of $f'(x)$ depends solely on $(x-1)$.
For strictly increasing, $f'(x) > 0 \implies x - 1 > 0 \implies x > 1$.
However, at $x=3$, $f'(x)=0$. So interval is $(1, 3) \cup (3, \infty)$.

Determine the intervals in which the function $f(x) = \sin^4 x + \cos^4 x$, where $x \in [0, \frac{\pi}{2}]$, is strictly increasing and strictly decreasing. Solution: $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x$.
$f'(x) = -\frac{1}{2} \cdot 2\sin 2x \cdot \cos 2x \cdot 2 = -\sin 4x$.
Strictly Increasing: $f'(x) > 0 \implies -\sin 4x > 0 \implies \sin 4x < 0$. For $x \in [0, \pi/2]$, $4x \in [0, 2\pi]$. $\sin 4x < 0$ in $(\pi, 2\pi) \implies \pi < 4x < 2\pi \implies x \in (\frac{\pi}{4}, \frac{\pi}{2})$.
Strictly Decreasing: $f'(x) < 0 \implies \sin 4x > 0 \implies 0 < 4x < \pi \implies x \in (0, \frac{\pi}{4})$.

Prove that the function $f(x) = \frac{4\sin x}{2 + \cos x} - x$ is an increasing function of $x$ in $[0, \frac{\pi}{2}]$. Solution: $f'(x) = \frac{(2+\cos x)(4\cos x) - (4\sin x)(-\sin x)}{(2+\cos x)^2} - 1 = \frac{8\cos x + 4\cos^2 x + 4\sin^2 x}{(2+\cos x)^2} - 1$.
$f'(x) = \frac{8\cos x + 4}{(2+\cos x)^2} - 1 = \frac{8\cos x + 4 - (4 + \cos^2 x + 4\cos x)}{(2+\cos x)^2} = \frac{4\cos x - \cos^2 x}{(2+\cos x)^2} = \frac{\cos x(4 - \cos x)}{(2+\cos x)^2}$.
In $[0, \pi/2]$, $\cos x \ge 0$ and $4 - \cos x > 0$. The denominator is always positive. Hence $f'(x) \ge 0$, proving it is an increasing function.

Find the equation of the tangent to the curve $y = \sqrt{3x-2}$ which is parallel to the line $4x - 2y + 5 = 0$. Solution: Slope of given line: $2y = 4x + 5 \implies y = 2x + 5/2 \implies m = 2$.
Derivative of curve: $\frac{dy}{dx} = \frac{3}{2\sqrt{3x-2}} = 2 \implies 3 = 4\sqrt{3x-2} \implies 9 = 16(3x-2) \implies 48x - 32 = 9 \implies 48x = 41 \implies x = \frac{41}{48}$.
Find $y$: $y = \sqrt{3(\frac{41}{48}) - 2} = \sqrt{\frac{41}{16} - \frac{32}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.
Equation of tangent: $y - \frac{3}{4} = 2(x - \frac{41}{48}) \implies y - \frac{3}{4} = 2x - \frac{41}{24} \implies 24y - 18 = 48x - 41 \implies 48x - 24y - 23 = 0$.

Find the points on the curve $y = x^3 - 2x^2 - x$ at which the tangent lines are parallel to the line $y = 3x - 2$. Solution: Slope of given line $m = 3$.
Curve derivative: $\frac{dy}{dx} = 3x^2 - 4x - 1 = 3 \implies 3x^2 - 4x - 4 = 0$.
Factor: $3x^2 - 6x + 2x - 4 = 0 \implies 3x(x-2) + 2(x-2) = 0 \implies (3x+2)(x-2) = 0 \implies x = 2, x = -2/3$.
If $x = 2$, $y = 8 - 8 - 2 = -2$. Point is $(2, -2)$.
If $x = -2/3$, $y = -8/27 - 8/9 + 2/3 = \frac{-8 - 24 + 18}{27} = -\frac{14}{27}$. Point is $(-2/3, -14/27)$.

Find the angle of intersection between the curves $y^2 = x$ and $x^2 = y$ at their non-zero point of intersection. Solution: Solving for intersection: $(x^2)^2 = x \implies x^4 - x = 0 \implies x(x^3 - 1) = 0$. Non-zero point is $(1, 1)$.
Curve 1 ($y^2=x$): $2y\frac{dy}{dx} = 1 \implies m_1 = \frac{1}{2y} \implies m_1 = 1/2$.
Curve 2 ($x^2=y$): $\frac{dy}{dx} = 2x \implies m_2 = 2$.
Angle $\tan\theta = |\frac{m_2 - m_1}{1 + m_1 m_2}| = |\frac{2 - 1/2}{1 + (2)(1/2)}| = \frac{3/2}{2} = \frac{3}{4}$.
Therefore, $\theta = \tan^{-1}(\frac{3}{4})$.

Find the equations of the normal to the curve $y = x^3 + 2x + 6$ which are parallel to the line $x + 14y + 4 = 0$. Solution: Slope of given line is $-1/14$. Since normal is parallel to this line, slope of normal $m_n = -1/14$.
Therefore, slope of tangent $m_t = 14$.
Curve derivative: $\frac{dy}{dx} = 3x^2 + 2 = 14 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2$.
When $x=2, y=8+4+6=18$. Normal: $y - 18 = -\frac{1}{14}(x - 2) \implies x + 14y - 254 = 0$.
When $x=-2, y=-8-4+6=-6$. Normal: $y + 6 = -\frac{1}{14}(x + 2) \implies x + 14y + 86 = 0$.

Use differentials to find the approximate value of $\sqrt{0.037}$, given that $\sqrt{0.04} = 0.2$. Solution: Let $y = \sqrt{x}$. Take $x = 0.04$ and $\Delta x = -0.003$.
$dy = \frac{1}{2\sqrt{x}}\Delta x = \frac{1}{2(0.2)}(-0.003) = \frac{-0.003}{0.4} = -0.0075$.
Approx value $= y + dy = 0.2 - 0.0075 = 0.1925$.

Find the approximate value of $f(5.001)$, where $f(x) = x^3 - 7x^2 + 15$. Solution: Let $x = 5$ and $\Delta x = 0.001$.
$f(5) = 125 - 7(25) + 15 = 125 - 175 + 15 = -35$.
$f'(x) = 3x^2 - 14x \implies f'(5) = 3(25) - 14(5) = 75 - 70 = 5$.
$\Delta y = f'(x)\Delta x = 5(0.001) = 0.005$.
Approx value $= -35 + 0.005 = -34.995$.

The radius of a sphere shrinks from $10\text{ cm}$ to $9.8\text{ cm}$. Find the approximate decrease in its volume using differentials. Solution: $V = \frac{4}{3}\pi r^3$. Given $r=10$ and $dr = -0.2$.
$dV = 4\pi r^2 dr = 4\pi (10^2) (-0.2) = 400\pi (-0.2) = -80\pi\text{ cm}^3$.
The approximate decrease in volume is $80\pi\text{ cm}^3$.

Find the local maxima and local minima for the function $f(x) = x^3 - 6x^2 + 9x + 15$. Solution: $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$. Setting $f'(x)=0 \implies x=1, x=3$.
$f''(x) = 6x - 12$.
At $x=1$: $f''(1) = -6 < 0 \implies$ Local Maxima. Max value $= 1 - 6 + 9 + 15 = 19$.
At $x=3$: $f''(3) = 6 > 0 \implies$ Local Minima. Min value $= 27 - 54 + 27 + 15 = 15$.

Find the absolute maximum and absolute minimum values of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the interval $[1, 5]$. Solution: $f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$. Critical points are $x=2$ and $x=3$ (both in interval).
Evaluate endpoints and critical points:
$f(1) = 2 - 15 + 36 + 1 = 24$
$f(2) = 16 - 60 + 72 + 1 = 29$
$f(3) = 54 - 135 + 108 + 1 = 28$
$f(5) = 250 - 375 + 180 + 1 = 56$
Absolute Max $= 56$ (at $x=5$). Absolute Min $= 24$ (at $x=1$).

Find the maximum and minimum values of the function $f(x) = x + \sin 2x$ on the interval $[0, 2\pi]$. Solution: $f'(x) = 1 + 2\cos 2x = 0 \implies \cos 2x = -1/2$.
In $[0, 2\pi]$, $2x \in [0, 4\pi]$. Solutions for $2x$ are $2\pi/3, 4\pi/3, 8\pi/3, 10\pi/3$.
So $x = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3$. Evaluate endpoints and critical points:
$f(0) = 0$. $f(2\pi) = 2\pi$.
$f(\pi/3) = \pi/3 + \sqrt{3}/2 \approx 1.91$
$f(2\pi/3) = 2\pi/3 - \sqrt{3}/2 \approx 1.23$
Absolute Max $= 2\pi$ (at $x=2\pi$). Absolute Min $= 0$ (at $x=0$).

Find the local maximum and local minimum values of $f(x) = \frac{x}{2} + \frac{2}{x}$ for $x > 0$. Solution: $f'(x) = \frac{1}{2} - \frac{2}{x^2} = 0 \implies \frac{1}{2} = \frac{2}{x^2} \implies x^2 = 4 \implies x = 2$ (since $x>0$).
$f''(x) = \frac{4}{x^3}$. At $x=2$, $f''(2) = 4/8 = 1/2 > 0$.
Hence, $x=2$ is a point of local minimum. Local min value $= 2/2 + 2/2 = 2$.

A wire of length $36\text{ m}$ is cut into two pieces. One piece is bent into a square and the other into an equilateral triangle. Find the lengths of the two pieces so that the combined area of the square and the triangle is minimum. Solution: Let length of square piece $= x$, triangle piece $= 36-x$. Side of square $a = x/4$, side of triangle $b = (36-x)/3$.
Total Area $A = (x/4)^2 + \frac{\sqrt{3}}{4}((36-x)/3)^2 = \frac{x^2}{16} + \frac{\sqrt{3}(36-x)^2}{36}$.
$A' = \frac{2x}{16} - \frac{2\sqrt{3}(36-x)}{36} = \frac{x}{8} - \frac{\sqrt{3}(36-x)}{18} = 0$.
$\frac{x}{8} = \frac{36\sqrt{3} - x\sqrt{3}}{18} \implies 18x = 288\sqrt{3} - 8\sqrt{3}x \implies x(18+8\sqrt{3}) = 288\sqrt{3}$.
$x = \frac{288\sqrt{3}}{18+8\sqrt{3}} = \frac{144\sqrt{3}}{9+4\sqrt{3}}$.
This is the length for the square. $A'' > 0$, confirming a minimum.

Find the dimensions of the maximum area rectangle that can be inscribed in a semicircle of radius $R$. Solution: Let the semicircle be $x^2 + y^2 = R^2, y \ge 0$. Let a vertex of the rectangle be $(x, y)$.
The base of the rectangle lies on the diameter, length $= 2x$, height $= y$.
Area $A = 2xy = 2x\sqrt{R^2 - x^2}$. Let $Z = A^2 = 4x^2(R^2 - x^2) = 4R^2 x^2 - 4x^4$.
$Z' = 8R^2 x - 16x^3 = 0 \implies 8x(R^2 - 2x^2) = 0 \implies x = R/\sqrt{2}$.
Then $y = \sqrt{R^2 - R^2/2} = R/\sqrt{2}$.
Dimensions: length $= 2x = \sqrt{2}R$, height $= R/\sqrt{2}$.

Show that the right circular cylinder of a given volume, which is open at the top, has a minimum total surface area when its height is equal to its base radius. Solution: Volume $V = \pi r^2 h \implies h = \frac{V}{\pi r^2}$.
Surface area (open top) $S = \pi r^2 + 2\pi r h = \pi r^2 + 2\pi r \left(\frac{V}{\pi r^2}\right) = \pi r^2 + \frac{2V}{r}$.
$S' = 2\pi r - \frac{2V}{r^2} = 0 \implies 2\pi r = \frac{2V}{r^2} \implies \pi r^3 = V$.
Substitute $V = \pi r^2 h \implies \pi r^3 = \pi r^2 h \implies r = h$.
$S'' = 2\pi + \frac{4V}{r^3} > 0$, so it is a minimum.

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $10\text{ m}$. Find the dimensions of the window to admit maximum light through the whole opening. Solution: Let width be $2x$ and height of rectangle be $y$. Radius of semicircle $= x$.
Perimeter $P = 2x + 2y + \pi x = 10 \implies y = 5 - x - \frac{\pi x}{2}$.
Area $A = (2x)(y) + \frac{1}{2}\pi x^2 = 2x\left(5 - x - \frac{\pi x}{2}\right) + \frac{1}{2}\pi x^2 = 10x - 2x^2 - \pi x^2 + \frac{1}{2}\pi x^2 = 10x - 2x^2 - \frac{\pi}{2}x^2$.
$A' = 10 - 4x - \pi x = 0 \implies x(4+\pi) = 10 \implies x = \frac{10}{\pi+4}$.
Width $2x = \frac{20}{\pi+4}$. Calculate $y = 5 - \frac{10}{\pi+4} - \frac{5\pi}{\pi+4} = \frac{5\pi+20-10-5\pi}{\pi+4} = \frac{10}{\pi+4}$.
So, width $= \frac{20}{\pi+4}\text{ m}$ and height $y = \frac{10}{\pi+4}\text{ m}$.

Show that the volume of the greatest cylinder which can be inscribed in a cone of height $h$ and semi-vertical angle $\alpha$ is $\frac{4}{27}\pi h^3 \tan^2\alpha$. Solution: Let cylinder have radius $r$ and height $h_1$. Radius of cone base $R = h\tan\alpha$.
By similar triangles, $\frac{h-h_1}{r} = \cot\alpha \implies h_1 = h - r\cot\alpha$.
Volume of cylinder $V = \pi r^2 h_1 = \pi r^2(h - r\cot\alpha) = \pi(hr^2 - r^3\cot\alpha)$.
$V' = \pi(2hr - 3r^2\cot\alpha) = 0 \implies 2h = 3r\cot\alpha \implies r = \frac{2h\tan\alpha}{3}$.
Max volume $V = \pi \left(\frac{4h^2\tan^2\alpha}{9}\right) \left(h - \frac{2h}{3}\right) = \pi \left(\frac{4h^2\tan^2\alpha}{9}\right) \left(\frac{h}{3}\right) = \frac{4}{27}\pi h^3 \tan^2\alpha$. Proved.