Miscellaneous Exercise Practice
Applications of Derivatives – Mixed Problems
Rate of Change
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Using differentials, find the approximate value of $(0.999)^{\frac{1}{10}}$. (Note: If approximations are removed from syllabus, treat this as a conceptual practice or skip).
Alternatively: A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lower most. Its semi-vertical angle is $\tan^{-1}(0.5)$. Water is poured into it at a constant rate of $5 \text{ m}^3/\text{hr}$. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is $4 \text{ m}$.
Alternatively: A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lower most. Its semi-vertical angle is $\tan^{-1}(0.5)$. Water is poured into it at a constant rate of $5 \text{ m}^3/\text{hr}$. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is $4 \text{ m}$.
Given: $\frac{dV}{dt} = 5$. Semi-vertical angle $\alpha = \tan^{-1}(0.5) \Rightarrow \tan \alpha = \frac{r}{h} = 0.5 = \frac{1}{2}$. So $r = \frac{h}{2}$.
Volume: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\frac{h}{2})^2 h = \frac{\pi}{12}h^3$.
Differentiate: $\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4}h^2 \frac{dh}{dt}$.
Substitute: $5 = \frac{\pi}{4}(4)^2 \frac{dh}{dt} = 4\pi \frac{dh}{dt}$.
Solve: $\frac{dh}{dt} = \frac{5}{4\pi} \text{ m/hr}$.
$\boxed{\frac{5}{4\pi} \text{ m/hr}}$
Increasing/Decreasing
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Show that the function $f$ given by $f(x) = \frac{\log x}{x}$ has a maximum at $x = e$.
Derivative: $f'(x) = \frac{x(\frac{1}{x}) - \log x(1)}{x^2} = \frac{1 - \log x}{x^2}$.
Critical Point: $f'(x) = 0 \Rightarrow 1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
Second Derivative: $f''(x) = \frac{x^2(-\frac{1}{x}) - (1-\log x)(2x)}{x^4} = \frac{-x - 2x(1-\log x)}{x^4} = \frac{-3 + 2\log x}{x^3}$.
Check Max: At $x=e$, $f''(e) = \frac{-3 + 2(1)}{e^3} = \frac{-1}{e^3} 0$.
Since $f''(e) 0$, $x=e$ is a point of maxima.
$\boxed{\text{Proven}}$
Maxima/Minima
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A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs ₹ 70 per sq. metre for the base and ₹ 45 per sq. metre for sides, what is the cost of least expensive tank?
Dimensions: Let length be $x$ and breadth be $y$. Depth $h=2$.
Volume $V = x \cdot y \cdot 2 = 8 \Rightarrow xy = 4 \Rightarrow y = 4/x$.
Volume $V = x \cdot y \cdot 2 = 8 \Rightarrow xy = 4 \Rightarrow y = 4/x$.
Area: Base $A_b = xy = 4$. Sides $A_s = 2(x+y)h = 4(x+y)$.
Cost Function $C$: $C = 70(xy) + 45[4(x+y)] = 70(4) + 180(x + \frac{4}{x}) = 280 + 180(x + \frac{4}{x})$.
Minimize C: $\frac{dC}{dx} = 180(1 - \frac{4}{x^2})$.
Set to 0: $1 - \frac{4}{x^2} = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2$.
Set to 0: $1 - \frac{4}{x^2} = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2$.
Calculate Cost: At $x=2$, $C = 280 + 180(2 + 2) = 280 + 180(4) = 280 + 720 = 1000$.
$\boxed{₹ 1000}$
Maxima/Minima
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The sum of the perimeter of a circle and square is $k$, where $k$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
Given: $2\pi r + 4x = k \Rightarrow x = \frac{k - 2\pi r}{4}$.
Area Sum $A$: $A = \pi r^2 + x^2 = \pi r^2 + \frac{(k-2\pi r)^2}{16}$.
Derivative: $\frac{dA}{dr} = 2\pi r + \frac{2(k-2\pi r)(-2\pi)}{16} = 2\pi r - \frac{\pi(k-2\pi r)}{4}$.
Critical Point: $2\pi r = \frac{\pi(k-2\pi r)}{4} \Rightarrow 8r = k - 2\pi r \Rightarrow k = 2r(4+\pi)$.
Substitute k: $4x = k - 2\pi r = 2r(4+\pi) - 2\pi r = 8r + 2\pi r - 2\pi r = 8r$.
$4x = 8r \Rightarrow x = 2r$.
$4x = 8r \Rightarrow x = 2r$.
Conclusion: Side of square = Diameter of circle.
$\boxed{\text{Proven}}$
Increasing/Decreasing
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Find the intervals in which the function $f$ given by $f(x) = \frac{4\sin x - 2x - x\cos x}{2+\cos x}$ is increasing or decreasing.
Simplify f(x): $f(x) = \frac{4\sin x}{2+\cos x} - x$.
Derivative: $f'(x) = \frac{(2+\cos x)(4\cos x) - 4\sin x(-\sin x)}{(2+\cos x)^2} - 1$
$= \frac{8\cos x + 4\cos^2 x + 4\sin^2 x}{(2+\cos x)^2} - 1 = \frac{8\cos x + 4}{(2+\cos x)^2} - 1$.
$= \frac{8\cos x + 4\cos^2 x + 4\sin^2 x}{(2+\cos x)^2} - 1 = \frac{8\cos x + 4}{(2+\cos x)^2} - 1$.
Simplify $f'(x)$: $\frac{8\cos x + 4 - (4 + \cos^2 x + 4\cos x)}{(2+\cos x)^2} = \frac{4\cos x - \cos^2 x}{(2+\cos x)^2} = \frac{\cos x(4-\cos x)}{(2+\cos x)^2}$.
Analysis: Since $4-\cos x 0$ and denominator $0$, sign depends on $\cos x$.
Increasing: $\cos x 0 \Rightarrow x \in (0, \pi/2) \cup (3\pi/2, 2\pi)$.
Decreasing: $\cos x 0 \Rightarrow x \in (\pi/2, 3\pi/2)$.
$\boxed{\text{Inc: } (0, \pi/2), \text{ Dec: } (\pi/2, 3\pi/2)}$
Maxima/Minima
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A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Setup: Let radius be $r$. Width of rectangle $2r$, height $x$.
Perimeter $P = 2r + 2x + \pi r = 10 \Rightarrow 2x = 10 - (2+\pi)r \Rightarrow x = 5 - (1+\frac{\pi}{2})r$.
Perimeter $P = 2r + 2x + \pi r = 10 \Rightarrow 2x = 10 - (2+\pi)r \Rightarrow x = 5 - (1+\frac{\pi}{2})r$.
Area A: $A = 2rx + \frac{1}{2}\pi r^2 = r(10 - (2+\pi)r) + \frac{1}{2}\pi r^2 = 10r - 2r^2 - \pi r^2 + \frac{1}{2}\pi r^2 = 10r - 2r^2 - \frac{\pi}{2}r^2$.
Derivative: $\frac{dA}{dr} = 10 - 4r - \pi r$.
Set to 0: $r(4+\pi) = 10 \Rightarrow r = \frac{10}{4+\pi}$.
Set to 0: $r(4+\pi) = 10 \Rightarrow r = \frac{10}{4+\pi}$.
Dimensions: Width $2r = \frac{20}{4+\pi}$, Height $x = \frac{10}{4+\pi}$.
$\boxed{\text{Width: } \frac{20}{4+\pi}, \text{ Height: } \frac{10}{4+\pi}}$