Mathematics Week 9 Graded Assignment
Graded Assignment
1. Match Functions with Graphs and Area
Question: Match the given functions with their graphs and the area under the curve over $[-1,1]$:
Function | Graph | Area under curve $[-1,1]$ |
---|---|---|
(i) $f(x) = x$ | (b) | $\int_{-1}^{1} x,dx = 0$ |
(ii) $f(x) = | x | $ |
(iii) $f(x) = x^2$ | (a) | $\int_{-1}^{1} x^2,dx = \frac{2}{3}$ |
Solution:
- (i) $\rightarrow$ (b) $\rightarrow$ (3) (Note: The “3” here is likely a typo or mislabel; area is 0, not “3”. The correct area solution is as above.)
- Corrected: (i) $\rightarrow$ (b) $\rightarrow$ area $= 0$
- **(ii) $\rightarrow$ (c) $\rightarrow$ area $= 1$
- **(iii) $\rightarrow$ (a) $\rightarrow$ area $= \frac{2}{3}$
2. Curves Enclosing Negative Area
Question: Which of the following curves enclose a negative area on the x-axis in the interval $1$? Area above the x-axis is positive, below is negative. If the area below the x-axis is more than above, the net area is negative.
Solution: Curve 2 and Curve 4 enclose a negative area.
3. Cylinder Inscribed in a Circle
Question: A cylinder of radius $x$ and height $2h$ is inscribed in a circle of radius $R$. Find the values of $h$ for maximum volume and maximum surface area.
Solution:
- Volume:
- From right triangle: $h^2 + x^2 = R^2$
- Volume $V = 2\pi x^2 h = 2\pi (R^2 - h^2) h = 2\pi R^2 h - 2\pi h^3$
- $\frac{dV}{dh} = 2\pi R^2 - 6\pi h^2$
- $\frac{dV}{dh} = 0 \Rightarrow h^2 = \frac{R^2}{3} \Rightarrow h = \frac{R}{\sqrt{3}}$
- $\frac{d^2V}{dh^2} = -12\pi h < 0$ $\Rightarrow$ maximum at $h = \frac{R}{\sqrt{3}}$
- Surface Area:
- Surface area $S = 4\pi x h = 4\pi h \sqrt{R^2 - h^2}$
- $\frac{dS}{dh} = 4\pi \left[\sqrt{R^2 - h^2} - \frac{h^2}{\sqrt{R^2 - h^2}}\right]$
- $\frac{dS}{dh} = 0 \Rightarrow R^2 - h^2 = h^2 \Rightarrow h^2 = \frac{R^2}{2} \Rightarrow h = \frac{R}{\sqrt{2}}$
- $\frac{d^2S}{dh^2}\left(\frac{R}{\sqrt{2}}\right) = -20\pi < 0$ $\Rightarrow$ maximum at $h = \frac{R}{\sqrt{2}}$
4. Minimum Difference of Functions
Question: Given $f_2(x) = x$ and $g_2(x) = e^x$ are increasing functions, find the minimum value of the difference on $1$.
Solution:
- Minimum is at $x = 0$:
- $f_2(0) = 0$, $g_2(0) = 1$
- Difference: $1$
5. Error in Prediction (Area Between Curves)
Question: Compute the error in prediction for company A (difference in area between $f_1(x) = x^3$ and $g_1(x) = \sqrt{x}$) and company B (difference in area between $f_2(x) = x$ and $g_2(x) = e^x$) over $1$.
Solution:
- Company A:
- $\left|\int_{0}^{1} (x^3 - \sqrt{x}),dx\right| = \left|\frac{1}{4} - \frac{2}{3}\right| = \frac{5}{12}$
- Company B:
- $\left|\int_{0}^{1} (x - e^x),dx\right| = \left|\frac{1}{2} - (e - 1)\right| = e - \frac{3}{2}$
- Conclusion:
- $e - \frac{3}{2} > \frac{5}{12}$
- Error for company B is greater than for company A.
6. Critical Points and Extrema
Question: Given $f(x) = x^3 - 3x + 1$, find critical points and classify them.
Solution:
- Derivative: $f’(x) = 3x^2 - 3$
- Critical points: $f’(x) = 0 \Rightarrow x = \pm 1$
- Second derivative: $f’’(x) = 6x$
- $f’’(1) = 6 > 0$ $\Rightarrow$ local minimum at $x=1$, $f(1) = -1$
- $f’’(-1) = -6 < 0$ $\Rightarrow$ local maximum at $x=-1$
7. Riemann Sum
Question: Given $f(x) = 2x^2 + \frac{5}{6}$ on $2$, divide into 3 subintervals and compute the Riemann sum using left endpoints.
Solution:
- Subintervals: $[0,2)$, $[2,4)$, $[4,6)$
- Left endpoints: $x = 0, 2, 4$
- Riemann sum:
- $2f(0) + 2f(2) + 2f(4)$
- $= 2\left[\frac{5}{6} + (8 + \frac{5}{6}) + (32 + \frac{5}{6})\right] = 2[40 + \frac{5}{2}] = 85$
8. Piecewise Function and Critical Points
Question: Given
$$ f(x) = \begin{cases} -2x + 3, & 0 \leq x \leq 10 \\ x^2, & 10 < x < 20 \end{cases} $$find critical points and global minimum.
Solution:
- Derivative:
- $f’(x) = -2$ for $0 < x < 10$
- $f’(x) = 2x$ for $10 < x < 20$
- Not differentiable at $x=10$
- Critical point: $x=10$
- Values:
- $f(0) = 3$
- $f(10) = -17$
- $f(20) = 400$
- Global minimum:
- Attained at $x=10$, value $= -17$
9. Optimization: Rectangle
Question: Given $x - y = 5$, find the area of rectangle $A = 2xy$ and its minimum.
Solution:
- Express $y$: $y = x - 5$
- Area: $A = 2x(x-5) = 2x^2 - 10x$
- Derivative: $A’(x) = 4x - 10$
- Critical point: $A’(x) = 0 \Rightarrow x = \frac{5}{2}$
- Second derivative: $A’’(x) = 4 > 0$ $\Rightarrow$ local minimum
- Minimum value:
- $A\left(\frac{5}{2}\right) = 2\left(\frac{25}{4}\right) - 10\left(\frac{5}{2}\right) = \frac{25}{2} - 25 = -\frac{25}{2}$
These are all the questions and solutions from the Week-9.pdf file1.