MATH 130 -- Tutorial Exercises -- Solutions

1999 D1 & E1, Mathematics IE

Week 10

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Calculus

Answer 1         (Question)

Calculate the derivative of each of the following functions:

(i)

For $f(x) = \sqrt[4]{\cos\bigl(x^3\bigr)}$ we have that $$\frac{\dD f}{\dD x} = \frac14\bigl(\cos(x^3)\bigr)^{-\frac34}\big... ...x^2) = {-}\frac34 x^2 \frac{\sin(x^3)\cos(x^3)}{\sqrt[4]{\cos\bigl(x^3\bigr)}}$$.

(ii)

For $f(x) = \displaystyle\frac{\log x}{x^2}$ we have that $$\frac{\dD f}{\dD x} = \frac{\frac1{x}x^2 - 2x\log x}{(x^2)^2} = \frac{1 - 2\log x}{x^3}$$.

(iii)

For $f(x) = \displaystyle\frac{e^x}{x^3}$ $$\frac{\dD f}{\dD x} = \frac{e^x x^3 - 3x^2 e^x}{(x^3)^2} = \frac{x-3}{x^4}e^x$$.

Answer 2         (Question)

(i)

The stationary points of $f(x) = \displaystyle\frac{\log x}{x^2}$ occur when $0 = \displaystyle\frac{\dD f}{\dD x} = \frac{1 - 2\log x}{x^3}$, using an answer from the previous question.
Thus the stationary point occurs at $(e^{\frac12}, \frac1{2e})\approx (1.649,0.184)$.

(ii)

The stationary points of $f(x) = x^2\log x$ occur when $0 = \displaystyle\frac{\dD f}{\dD x} = 2x\log x + x = x(1+2\log x)$.
The only solution is when $x=e^{{-}\frac12}$ so the stationary point for $f$ is at $(e^{{-}\frac12},{-}\frac12 e^{-1})\approx (0.607,{-}0.184)$.
(Note that $x=0$ is outside the domain for $f$, since $\log x$ is not defined at $x=0$.)

(iii)

The stationary points of $f(x) = x^2 e^x$ occur when $0 = \displaystyle\frac{\dD f}{\dD x} = (x^2+2x)e^x$. Since $e^x > 0$ for all $x$, the only solutions are when $x=0$ and when $x={-}2$.
The stationary points for $f$ are at $(0,0)$ and $({-2},4e^{-2})$.

Algebra

Answer 1         (Question)

For the sum of odd integers $1+3+5+\dots +999$, use the formula $\frac{1}{2}n(a_1+a_n)$, where $a_1$ is the 1st summand, and $a_n$ is the last summand, where there are $n$ summands in all; in this case 500.
This gives an answer of: $\frac12\times 500(1+999) = 250,000$.

Answer 2         (Question)

Given the 7th term $a_7=6$ and the 14th term $a_{14}={-}15$, of an arithmetic progression, then the difference $d$ between successive terms satifies ${-}21 = {-}15 - 6 = d(14-7)$, hence $d={-3}$.
Now the 10th term is calculated from $a_{10}-a_7 = d(10-7) = {-}9$, so that $a_{10} = 6-9 = {-}3$.
(Check also that $a_{14}-a_{10} = -{15} - ({-}3) = {-}12 = 4d$, as required.)

Answer 3         (Question)

Over the first 10 years, Jack earns $10000 + 11000 + 12100 + \dots + 10000(1.1)^9$ in dollars.
This totals $10000\bigl((1.1)^{10}-1\bigr)/(1.1 - 1) = 159374$.

Over the first 10 years, Jill earns $20000 + 20500 + 21000 + \dots + 24500$ in dollars.
This totals $\frac12 \, 10(20000 + 24500) = 222500$, which is more than Jack has earned.

Note, however, that in the 11th year Jack earns $25937, which is greater than Jill's $25000.
Furthermore, Jack's increases are now at least $2500 per year, while Jill's increases remain at $500, so Jack will be earning significantly more than Jill in these later years.

Answer 4         (Question)

The ratio of successive terms in the G.P. is $r = \frac{1}{\sqrt{2}}<1$.
Hence the sum to infinity exists, and is given by $S_{\infty} = 2 \displaystyle\frac{1}{1-\frac{1}{\sqrt{2}}} = 2\frac{\sqrt{2}}{\sqrt{2}-1} = 2\sqrt{2}(\sqrt{2}+1) = 4 - 2\sqrt{2}$.

Answer 5         (Question)

For the G.P.: $2,\displaystyle\frac{2}{\sqrt{3}-3},\frac{1}{6-3\sqrt{3}},\,\dots\;$ the ratio of successive terms is $r=\displaystyle\frac{1}{\sqrt{3}-3}$ which satisfies $\bigl\vert r\bigr\vert<1$ since $\sqrt{3}\approx 1.732\dots\;$.

Thus the sum over infinitely many terms exists, and is given by $S_{\infty} = 2\frac{1}{1-\frac{1}{\sqrt{3}-3}} = 2\frac{\sqrt{3}-3}{\sqrt{3}-4}$.
This can be simplified to $\frac{2}{13}(3-\sqrt{3})(\sqrt{3}+4)= \frac{2}{13}(9-\sqrt{3})$.

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