Mathematics Week 9 Graded Assignment

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 $[-1,1]$:

FunctionGraphArea 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.