Need help, URGENT maths question: (4 Viewers)

1008 · Mar 10, 2016

Does anyone have any ideas on how to do the following question, I'm stumped...

InteGrand · Mar 10, 2016

1008 said:
Does anyone have any ideas on how to do the following question, I'm stumped...

View attachment 32990

$\noindent Let $a = \beta + \gamma - \alpha$, $b = \gamma + \alpha - \beta$, and $c = \alpha + \beta -\gamma$. Then $a,b,c>0$ due to the inequality conditions on $\alpha, \beta, \gamma$. It also easy to check that these are all unequal. For instance, suppose $a=b$. Then $\beta +\gamma - \alpha = \gamma + \alpha - \beta \Rightarrow \beta - \alpha = \alpha - \beta \Rightarrow \beta - \alpha = 0 \Rightarrow \beta = \alpha$, a contradiction. Now, from the given result, we have$

$\begin{align*}\left(\beta + \gamma - \alpha\right) \left( \gamma + \alpha - \beta\right) \left( \alpha + \beta -\gamma\right) = abc &<\frac{1}{8}\left(b+c\right) \left(c+a\right)\left(a+b\right) \\ &= \frac{1}{8}\left(2\alpha\right) \left(2\beta\right)\left(2\gamma\right) \quad (\text{by our definitions of }a,b,c) \\ &= \alpha \beta \gamma, \end{align*}$

$\noindent as required. Q.E.D.$

4025808 · Mar 10, 2016

Ah, I remember this question from the MATH1151 first class test

1008 · Mar 11, 2016

Thanks InteGrand, got a calculus test coming up from next week, so you're a life saver!

BTW 4025808 yeah, it's from the Calculus Problems booklet that you receive

And any of you know how to solve this one:
A lighthouse containing a revolving beacon is located 3km from P, the nearest point on a straight shoreline. The beacon revolves with a constant rotation rate of 4 revolutions per minute and throws a spot of light onto the shoreline. How fast is the spot of light moving when it is (a) at P and (b) at a point on the shoreline 2km from P?

InteGrand · Mar 11, 2016

1008 said:
Thanks InteGrand, got a calculus test coming up from next week, so you're a life saver!

BTW 4025808 yeah, it's from the Calculus Problems booklet that you receive

And any of you know how to solve this one:
A lighthouse containing a revolving beacon is located 3km from P, the nearest point on a straight shoreline. The beacon revolves with a constant rotation rate of 4 revolutions per minute and throws a spot of light onto the shoreline. How fast is the spot of light moving when it is (a) at P and (b) at a point on the shoreline 2km from P?

$\noindent Note that the angular velocity of the beacon in radians per second is $\omega = 4 \times \frac{2\pi}{60} = \frac{2}{15}\pi$. Now, clearly, when the light beacon is right at $P$, the rate (speed) at which it moves along the shoreline is just $v=r\omega = 3\times \frac{2}{15}\pi = \frac{2}{5}\pi\text{ km/s}$.$

$\noindent In the general case let the lighthouse be at point $L$, and let $P$ be directly above $L$ (so $PL = 3$, where distances are in km). Say that the beacon is rotating clockwise (so the beam of light goes from left to right along the shoreline on the page). Consider a point $X$ that is $x$ km right of $P$ on the shoreline (if $x<0$, then $X$ is actually left of $P$ by $-x$ metres) and say the light beacon is currently shining on that point. Let $\angle PLX = \theta$ (note that if $x>0$, we set $\theta >0$, and if $x<0$, then $\theta <0$. Also, since light is shining on the shoreline, the light beacon must be above the horizontal through $L$, so we have $-\frac{\pi}{2}<\theta < \frac{\pi}{2}$; if $\theta$ is outside this range, then the beacon is actually pointing down and not casting light on the shoreline).$

$\noindent Since $\angle LPX$ is a right angle, we have $x=3\tan \theta$ by basic trigonometry. Note that $\tan \theta = \frac{x}{3} \Rightarrow \sec^2 \theta = \frac{x^2}{9} +1$. Now, the rate at which the light beacon is moving along the shore is$

$\begin{align*}\frac{\mathrm{d}x}{\mathrm{d}t}&= 3\sec^2 \theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}t} \quad (\text{where }t \text{ is time measured in seconds, and we have used the chain rule}) \\ &= 3\left(\frac{x^2}{9} +1\right)\cdot \frac{\mathrm{d}\theta}{\mathrm{d}t} \\ &= \left(\frac{x^2}{3} +3\right)\cdot \frac{\mathrm{d}\theta}{\mathrm{d}t}. \\ &= \left(\frac{x^2}{3} +3\right)\cdot \frac{2}{15}\pi \quad (\text{as we are given the value of the (constant) angular velocity, which we converted to radians per second above})\end{align*}$

$\noindent Now, when the distance of the point in question from $P$ is 2 km, we must have $x^2 = 4$, so at this time, the rate of movement of the light is$

$\begin{align*}\frac{\mathrm{d}x}{\mathrm{d}t} \Big{|}_{t=2} &= \left(\frac{4}{3}+3\right)\cdot \frac{2}{15}\pi \\ &= \frac{13}{3}\times \frac{2}{15}\pi \\ &= \frac{26}{45}\pi \text{ km/s}. \end{align*}$

$\noindent Note that our general formula for $\frac{\mathrm{d}x}{\mathrm{d}t}$ also works for the $x=0$ case we did at the start (matching with our answer we gave for it).$

1008 · Mar 12, 2016

InteGrand said:
$\noindent Note that the angular velocity of the beacon in radians per second is $\omega = 4 \times \frac{2\pi}{60} = \frac{2}{15}\pi$. Now, clearly, when the light beacon is right at $P$, the rate (speed) at which it moves along the shoreline is just $v=r\omega = 3\times \frac{2}{15}\pi = \frac{2}{5}\pi\text{ km/s}$.$

$\noindent In the general case let the lighthouse be at point $L$, and let $P$ be directly above $L$ (so $PL = 3$, where distances are in km). Say that the beacon is rotating clockwise (so the beam of light goes from left to right along the shoreline on the page). Consider a point $X$ that is $x$ km right of $P$ on the shoreline (if $x<0$, then $X$ is actually left of $P$ by $-x$ metres) and say the light beacon is currently shining on that point. Let $\angle PLX = \theta$ (note that if $x>0$, we set $\theta >0$, and if $x<0$, then $\theta <0$. Also, since light is shining on the shoreline, the light beacon must be above the horizontal through $L$, so we have $-\frac{\pi}{2}<\theta < \frac{\pi}{2}$; if $\theta$ is outside this range, then the beacon is actually pointing down and not casting light on the shoreline).$

$\noindent Since $\angle LPX$ is a right angle, we have $x=3\tan \theta$ by basic trigonometry. Note that $\tan \theta = \frac{x}{3} \Rightarrow \sec^2 \theta = \frac{x^2}{9} +1$. Now, the rate at which the light beacon is moving along the shore is$

$\begin{align*}\frac{\mathrm{d}x}{\mathrm{d}t}&= 3\sec^2 \theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}t} \quad (\text{where }t \text{ is time measured in seconds, and we have used the chain rule}) \\ &= 3\left(\frac{x^2}{9} +1\right)\cdot \frac{\mathrm{d}\theta}{\mathrm{d}t} \\ &= \left(\frac{x^2}{3} +3\right)\cdot \frac{\mathrm{d}\theta}{\mathrm{d}t}. \\ &= \left(\frac{x^2}{3} +3\right)\cdot \frac{2}{15}\pi \quad (\text{as we are given the value of the (constant) angular velocity, which we converted to radians per second above})\end{align*}$

$\noindent Now, when the distance of the point in question from $P$ is 2 km, we must have $x^2 = 4$, so at this time, the rate of movement of the light is$

$\begin{align*}\frac{\mathrm{d}x}{\mathrm{d}t} \Big{|}_{t=2} &= \left(\frac{4}{3}+3\right)\cdot \frac{2}{15}\pi \\ &= \frac{13}{3}\times \frac{2}{15}\pi \\ &= \frac{26}{45}\pi \text{ km/s}. \end{align*}$

$\noindent Note that our general formula for $\frac{\mathrm{d}x}{\mathrm{d}t}$ also works for the $x=0$ case we did at the start (matching with our answer we gave for it).$

Thanks again InteGrand. I have another question :

To prove that g is the inverse function to f and not vice versa, do I need to state that "the domain of f --> [0,inf) doesn't fully lie in the domain of g --> [1,inf), so g is the inverse of f"? This is using the hint that they've given at the end of the question where it states "remember (f o g)(x) = f(g(x)) whenever x is in the domain of g and g(x) is in the domain of f"

InteGrand · Mar 12, 2016

1008 said:
Thanks again InteGrand. I have another question :

To prove that g is the inverse function to f and not vice versa, do I need to state that "the domain of f --> [0,inf) doesn't fully lie in the domain of g --> [1,inf), so g is the inverse of f"? This is using the hint that they've given at the end of the question where it states "remember (f o g)(x) = f(g(x)) whenever x is in the domain of g and g(x) is in the domain of f"

$\noindent In general, for any functions $f,g$, if $g$ is the inverse of $f$, then $f$ is also necessarily the inverse of $g$. In this particular case, $f$'s domain is the codomain of $g$ and $f$'s codomain is the domain of $g$. This is something necessary for two functions to be inverses. (It's not important whether the domain of $f$ is contained in the domain of $g$.)$

1008 · Mar 14, 2016

InteGrand said:
$\noindent In general, for any functions $f,g$, if $g$ is the inverse of $f$, then $f$ is also necessarily the inverse of $g$. In this particular case, $f$'s domain is the codomain of $g$ and $f$'s codomain is the domain of $g$. This is something necessary for two functions to be inverses. (It's not important whether the domain of $f$ is contained in the domain of $g$.)$

Alright thanks, I've repped you, but I don't think it'll make much difference. I do have a few more questions though, if InteGrand or someone else would like to help:

InteGrand · Mar 14, 2016

1008 said:
Alright thanks, I've repped you, but I don't think it'll make much difference. I do have a few more questions though, if InteGrand or someone else would like to help:

$\noindent 1. This is just the interior of the triangular region bounded by the lines $y=2x$, $x=0$ and $x=2$, as well as the boundaries. The sketch is in the link below:$

http://www.wolframalpha.com/input/?i=0<=+y+<=+2x+and+0<=+x+<=+2.

$\noindent 2. This turns out to be the same as the region in Q.1. To see why this is the case, observe that the condition $\frac{y}{2}\leq x \leq 2$ is equivalent to $y\leq 2x \leq 4$. So we know $y\leq 2x$. We also know that $0\leq y \leq 4$ and $x\leq 2$. Moreover, $x\geq \frac{y}{2} \Rightarrow x \geq 0$. So $0\leq x \leq 2$. So the region has $y\leq 2x$ and $0\leq x \leq 2$, as in Q.1. Here's the plot again (with the relations given in Q.2):$

http://www.wolframalpha.com/input/?i=y/2+<=+x+<=+2+and+0<=+y+<=+4 .

$\noindent 3. For $f(x) = \sqrt{1-2\sin x}$, the maximal domain would have $1-2\sin x\geq 0 \Longleftrightarrow \sin x \leq \frac{1}{2}$. Now, from a sketch of the graph of $y=\sin x$, it is clear that one place where this happens is in the interval where $-\frac{7\pi}{6} \leq x \leq \frac{\pi}{6}$. Now, we can shift this interval in either direction (left or right) by a multiple of $2\pi$ and the inequation will still hold (due to the $2\pi$ periodicity of the sine function). Hence the required domain may be written as: $\left \{x \in \mathbb{R}: -\frac{7\pi}{6}+2k\pi \leq x \leq \frac{\pi}{6}+2k\pi, \, k\in \mathbb{Z}\right \}$.$

$\noindent For the second function they gave, the maximal domain is just $\mathbb{R}$, since the function is well-defined at any real number.$

$\noindent 4. (b) I believe that the question as typed has a typo. Instead of $v(t) = 50 -\left(1-\mathrm{e}^{-\frac{t}{5}}\right)$, it should clearly be $v(t) = 50\left(1-\mathrm{e}^{-\frac{t}{5}}\right)$. This is so that $v(0)=0$ instead of $v(0)=50$, $\Big{(}$which would be strange, since the parcel is just dropped, not hurled out at 50 metres per second. Furthermore, the answer to the question would actually be right from $t=0$ if the function were $v(t) = 50 -\left(1-\mathrm{e}^{-\frac{t}{5}}\right)$, which would have made it an odd question.\Big{)}. Now, the limiting velocity is $v_T = 50$. Note that $v(t) < v_T$ for all $t\geq 0$. So we want to find the $t$ such that $v_T-v(t)=1$. To do this, we just solve for $t$:$

$\begin{align*} v_T- v(t) &=1 \Longleftrightarrow 50 - 50\left(1-\mathrm{e}^{-\frac{t}{5}}\right) &= 1 \\ \Longleftrightarrow 50\left(1-\mathrm{e}^{-\frac{t}{5}}\right) &= 49 \\ \Longleftrightarrow 1-\mathrm{e}^{-\frac{t}{5}} = \frac{49}{50} \\ \Longleftrightarrow \mathrm{e}^{-\frac{t}{5}} &= \frac{1}{50} \\ \Longleftrightarrow \frac{t}{5} &= \ln 50 \\ \Longleftrightarrow t &= 5 \ln 50 = 19.56011\ldots. \end{align*}$

$\noindent So the answer is after $5\ln 50$ seconds.$

$\noindent 5. Observe that as $x\to \infty$, $\frac{1}{x}\to 0$, so $\cos \left(\frac{1}{x}\right) \to \cos 0 =1$, as the cosine function is continuous.$

$\noindent 6. The $\mathrm{sech} ^{-1}$ function is obtained by flipping the right-half of the graph of $y=\mathrm{sech}\, x$ about the line $y=x$. The domain of the $\mathrm{sech}^{-1}$ function is $(0,1]$ (there is a vertical asymptote at $x=0$). The graph is linked below:$

http://www.wolframalpha.com/input/?i=sech^(-1)(x)&rawformassumption=&lk=1 .

Zoinked · Mar 14, 2016

why integrand isnt a maths mod is beyond me

1008 · Mar 15, 2016

InteGrand said:
$\noindent 1. This is just the interior of the triangular region bounded by the lines $y=2x$, $x=0$ and $x=2$, as well as the boundaries. The sketch is in the link below:$

http://www.wolframalpha.com/input/?i=0<=+y+<=+2x+and+0<=+x+<=+2.

$\noindent 2. This turns out to be the same as the region in Q.1. To see why this is the case, observe that the condition $\frac{y}{2}\leq x \leq 2$ is equivalent to $y\leq 2x \leq 4$. So we know $y\leq 2x$. We also know that $0\leq y \leq 4$ and $x\leq 2$. Moreover, $x\geq \frac{y}{2} \Rightarrow x \geq 0$. So $0\leq x \leq 2$. So the region has $y\leq 2x$ and $0\leq x \leq 2$, as in Q.1. Here's the plot again (with the relations given in Q.2):$

http://www.wolframalpha.com/input/?i=y/2+<=+x+<=+2+and+0<=+y+<=+4 .

$\noindent 3. For $f(x) = \sqrt{1-2\sin x}$, the maximal domain would have $1-2\sin x\geq 0 \Longleftrightarrow \sin x \leq \frac{1}{2}$. Now, from a sketch of the graph of $y=\sin x$, it is clear that one place where this happens is in the interval where $-\frac{7\pi}{6} \leq x \leq \frac{\pi}{6}$. Now, we can shift this interval in either direction (left or right) by a multiple of $2\pi$ and the inequation will still hold (due to the $2\pi$ periodicity of the sine function). Hence the required domain may be written as: $\left \{x \in \mathbb{R}: -\frac{7\pi}{6}+2k\pi \leq x \leq \frac{\pi}{6}+2k\pi, \, k\in \mathbb{Z}\right \}$.$

$\noindent For the second function they gave, the maximal domain is just $\mathbb{R}$, since the function is well-defined at any real number.$

$\noindent 4. (b) I believe that the question as typed has a typo. Instead of $v(t) = 50 -\left(1-\mathrm{e}^{-\frac{t}{5}}\right)$, it should clearly be $v(t) = 50\left(1-\mathrm{e}^{-\frac{t}{5}}\right)$. This is so that $v(0)=0$ instead of $v(0)=50$, $\Big{(}$which would be strange, since the parcel is just dropped, not hurled out at 50 metres per second. Furthermore, the answer to the question would actually be right from $t=0$ if the function were $v(t) = 50 -\left(1-\mathrm{e}^{-\frac{t}{5}}\right)$, which would have made it an odd question.\Big{)}. Now, the limiting velocity is $v_T = 50$. Note that $v(t) < v_T$ for all $t\geq 0$. So we want to find the $t$ such that $v_T-v(t)=1$. To do this, we just solve for $t$:$

$\begin{align*} v_T- v(t) &=1 \Longleftrightarrow 50 - 50\left(1-\mathrm{e}^{-\frac{t}{5}}\right) &= 1 \\ \Longleftrightarrow 50\left(1-\mathrm{e}^{-\frac{t}{5}}\right) &= 49 \\ \Longleftrightarrow 1-\mathrm{e}^{-\frac{t}{5}} = \frac{49}{50} \\ \Longleftrightarrow \mathrm{e}^{-\frac{t}{5}} &= \frac{1}{50} \\ \Longleftrightarrow \frac{t}{5} &= \ln 50 \\ \Longleftrightarrow t &= 5 \ln 50 = 19.56011\ldots. \end{align*}$

$\noindent So the answer is after $5\ln 50$ seconds.$

$\noindent 5. Observe that as $x\to \infty$, $\frac{1}{x}\to 0$, so $\cos \left(\frac{1}{x}\right) \to \cos 0 =1$, as the cosine function is continuous.$

$\noindent 6. The $\mathrm{sech} ^{-1}$ function is obtained by flipping the right-half of the graph of $y=\mathrm{sech}\, x$ about the line $y=x$. The domain of the $\mathrm{sech}^{-1}$ function is $(0,1]$ (there is a vertical asymptote at $x=0$). The graph is linked below:$

http://www.wolframalpha.com/input/?i=sech^(-1)(x)&rawformassumption=&lk=1 .

I think it's the first time I've seen you offline when I'm writing in this thread.

BTW, you were right about Question 4(b). I indeed did make a typo...thanks for picking that up. I still don't understand your explanation for question 5, as I believe the question asks for considering the limit of cos(1/x) as x approaches 0, not infinity...could you please explain it further?

Also, I've got more questions that I need help with. I know you're busy, but it would be great if you or someone else could please answer these:

Thanks in advance

InteGrand · Mar 15, 2016

1008 said:
I think it's the first time I've seen you offline when I'm writing in this thread.

BTW, you were right about Question 4(b). I indeed did make a typo...thanks for picking that up. I still don't understand your explanation for question 5, as I believe the question asks for considering the limit of cos(1/x) as x approaches 0, not infinity...could you please explain it further?

Also, I've got more questions that I need help with. I know you're busy, but it would be great if you or someone else could please answer these:

Thanks in advance

$\noindent Oh right, sorry, I misread it (Q5). In that case, since $\frac{1}{x}$ tends to $\pm \infty$ as $x \to 0^{\pm}$, $\cos \left(\frac{1}{x}\right)$ will just oscillate infinitely often near 0, so the limit does not exist.$

1008 · Mar 15, 2016

InteGrand said:
$\noindent Oh right, sorry, I misread it (Q5). In that case, since $\frac{1}{x}$ tends to $\pm \infty$ as $x \to 0^{\pm}$, $\cos \left(\frac{1}{x}\right)$ will just oscillate infinitely often near 0, so the limit does not exist.$

Thanks for your reply...Any thoughts on the other questions?

InteGrand · Mar 15, 2016

1008 said:
I think it's the first time I've seen you offline when I'm writing in this thread.

BTW, you were right about Question 4(b). I indeed did make a typo...thanks for picking that up. I still don't understand your explanation for question 5, as I believe the question asks for considering the limit of cos(1/x) as x approaches 0, not infinity...could you please explain it further?

Also, I've got more questions that I need help with. I know you're busy, but it would be great if you or someone else could please answer these:

Thanks in advance

$\noindent iii) (First question) (a) Recall that $\cos \xi \leq 1$ for all $\xi \in \mathbb{R}$. Let $x\geq 0$. Integrating both sides of the inequality from $0$ to $x$ gives us $\Big{[}\sin \xi \Big{]}_0 ^x \leq x \Rightarrow \sin x \leq x$.$

$\noindent (b) We showed above that $\sin \xi \leq \xi$ for all $\xi \geq 0$. Integrate both sides from $0$ to $x$ (where $x\geq 0$):$

$\begin{align*}\int _0 ^x \sin \xi \text{ d}\xi &\leq \int _0 ^x \xi \text{ d}\xi \\ \Rightarrow \Big{[}-\cos \xi \Big{]}_0 ^x &\leq x^2 \\ \Rightarrow -\cos x + \cos 0 & \leq x^2 \\ \Longleftrightarrow -\cos x + 1& \leq x^2 \\ \Longleftrightarrow \cos x \geq 1 -\frac{x^2}{2}.\end{align*}$

$\noindent (ii) (a) Given: $x^2 +xy+y^2 =9$. Using implicit differentiation (and the product rule where needed), $2x + y+xy^\prime + 2yy^\prime = 0 \Rightarrow 2x+y + \left(x + 2y\right)y^\prime = 0 \Rightarrow y^\prime = - \frac{2x+y}{x + 2y}$, assuming $x+2y \neq 0$.$

$\noindent (b) The tangents are horizontal when $y^\prime =0$. This happens only when $2x+y = 0 \Rightarrow y = -2x$. Sub. this into the curve's equation to find the points: $x^2 + x(-2x) + (-2x)^2 = 9 \Longleftrightarrow x^2 - 2x^2 +4x^2 = 9 \Longleftrightarrow 3x^2=9 \Longleftrightarrow x^2 = 3 \Longleftrightarrow x = \pm \sqrt{3}$. Since $y=-2x$, $y=\mp 2 \sqrt{3}$. So the desired points are the two points $\left(\pm \sqrt{3}, \mp 2 \sqrt{3}\right)$.$

$\noindent 2. (i) (a) Let $u\equiv u(x)=x^2 + x$. Then $\frac{\mathrm{d}u}{\mathrm{d}x}\equiv u^\prime (x)=2x +1$. Also, by the Fundamental Theorem of Calculus, $g^\prime (u) = \sqrt{1+u^4}=\sqrt{1+\left(x^2 + x\right)^4}$, where $g(u) = \int _2 ^u \sqrt{1+t^4}\text{ d}t$ (note that $f(x) = g(u(x))$). So, using the chain rule, $f^\prime (x) = g^\prime (u) \cdot u^\prime (x)\Rightarrow u^\prime (x) = \sqrt{1+\left(x^2 + x\right)^4}\cdot (2x +1)$.$

$\noindent (b) The stationary point is when $f^\prime (x)=0$, which happens when and only when $2x +1=\Longleftrightarrow x = -\frac{1}{2}$. The nature is that this is a local minimum, because clearly $f^\prime (x) <0$ just to the left of its zero, and positive just to the right (since it follows the sign of $2x+1$). So in summary, the stationary point is $-\frac{1}{2}$, and it is a local minimum.$

$\noindent (c) For $f$ to be invertible, we want its derivative to have the same sign. As we said before, the sign of $f^\prime (x)$ is just that of $2x+1$, since the other factor in $f^\prime (x)$ is always positive. So we can make it invertible on either of the following two intervals: $I_1 := \left(-\infty, -\frac{1}{2} \right]$, $I_2 := \left[-\frac{1}{2}, \infty \right )$. Since we want it to contain $0$, we take $I =I_2 =\left[-\frac{1}{2},\infty \right)$.$

$\noindent (d) Observe that $f(x) = 0$ iff $x^2 + x = 2$ (this is because the integrand of the integral definition of $f$ is of fixed sign, so it we integrate over a non-degenerate interval, the integral will be not be 0; but integrating over a degenerate interval always yields 0). In the domain $I$, this happens when and only when $x=1$ (solving the quadratic). So using the inverse function theorem, $\left(f^{-1}\right)^\prime (0) = \frac{1}{f^\prime (1)}=\frac{1}{3\sqrt{17}}$ (subbing in to the expression for $f^\prime (x)$).$

1008 · Mar 16, 2016

InteGrand said:
$\noindent iii) (First question) (a) Recall that $\cos \xi \leq 1$ for all $\xi \in \mathbb{R}$. Let $x\geq 0$. Integrating both sides of the inequality from $0$ to $x$ gives us $\Big{[}\sin \xi \Big{]}_0 ^x \leq x \Rightarrow \sin x \leq x$.$

$\noindent (b) We showed above that $\sin \xi \leq \xi$ for all $\xi \geq 0$. Integrate both sides from $0$ to $x$ (where $x\geq 0$):$

$\begin{align*}\int _0 ^x \sin \xi \text{ d}\xi &\leq \int _0 ^x \xi \text{ d}\xi \\ \Rightarrow \Big{[}-\cos \xi \Big{]}_0 ^x &\leq x^2 \\ \Rightarrow -\cos x + \cos 0 & \leq x^2 \\ \Longleftrightarrow -\cos x + 1& \leq x^2 \\ \Longleftrightarrow \cos x \geq 1 -\frac{x^2}{2}.\end{align*}$

$\noindent (ii) (a) Given: $x^2 +xy+y^2 =9$. Using implicit differentiation (and the product rule where needed), $2x + y+xy^\prime + 2yy^\prime = 0 \Rightarrow 2x+y + \left(x + 2y\right)y^\prime = 0 \Rightarrow y^\prime = - \frac{2x+y}{x + 2y}$, assuming $x+2y \neq 0$.$

$\noindent (b) The tangents are horizontal when $y^\prime =0$. This happens only when $2x+y = 0 \Rightarrow y = -2x$. Sub. this into the curve's equation to find the points: $x^2 + x(-2x) + (-2x)^2 = 9 \Longleftrightarrow x^2 - 2x^2 +4x^2 = 9 \Longleftrightarrow 3x^2=9 \Longleftrightarrow x^2 = 3 \Longleftrightarrow x = \pm \sqrt{3}$. Since $y=-2x$, $y=\mp 2 \sqrt{3}$. So the desired points are the two points $\left(\pm \sqrt{3}, \mp 2 \sqrt{3}\right)$.$

$\noindent 2. (i) (a) Let $u\equiv u(x)=x^2 + x$. Then $\frac{\mathrm{d}u}{\mathrm{d}x}\equiv u^\prime (x)=2x +1$. Also, by the Fundamental Theorem of Calculus, $g^\prime (u) = \sqrt{1+u^4}=\sqrt{1+\left(x^2 + x\right)^4}$, where $g(u) = \int _2 ^u \sqrt{1+t^4}\text{ d}t$ (note that $f(x) = g(u(x))$). So, using the chain rule, $f^\prime (x) = g^\prime (u) \cdot u^\prime (x)\Rightarrow u^\prime (x) = \sqrt{1+\left(x^2 + x\right)^4}\cdot (2x +1)$.$

$\noindent (b) The stationary point is when $f^\prime (x)=0$, which happens when and only when $2x +1=\Longleftrightarrow x = -\frac{1}{2}$. The nature is that this is a local minimum, because clearly $f^\prime (x) <0$ just to the left of its zero, and positive just to the right (since it follows the sign of $2x+1$). So in summary, the stationary point is $-\frac{1}{2}$, and it is a local minimum.$

$\noindent (c) For $f$ to be invertible, we want its derivative to have the same sign. As we said before, the sign of $f^\prime (x)$ is just that of $2x+1$, since the other factor in $f^\prime (x)$ is always positive. So we can make it invertible on either of the following two intervals: $I_1 := \left(-\infty, -\frac{1}{2} \right]$, $I_2 := \left[-\frac{1}{2}, \infty \right )$. Since we want it to contain $0$, we take $I =I_2 =\left[-\frac{1}{2},\infty \right)$.$

$\noindent (d) Observe that $f(x) = 0$ iff $x^2 + x = 2$ (this is because the integrand of the integral definition of $f$ is of fixed sign, so it we integrate over a non-degenerate interval, the integral will be not be 0; but integrating over a degenerate interval always yields 0). In the domain $I$, this happens when and only when $x=1$ (solving the quadratic). So using the inverse function theorem, $\left(f^{-1}\right)^\prime (0) = \frac{1}{f^\prime (1)}=\frac{1}{3\sqrt{17}}$ (subbing in to the expression for $f^\prime (x)$).$

Thanks again, InteGrand, you're helping me so much! Really appreciate it! Just got a few more questions, if you don't mind. Also outta curiosity, how do you type those mathematical symbols on BoS? AND HOW R U SUCH A GOD AT MATHS? I mean, WHAT DO YOU DO?

InteGrand · Mar 16, 2016

1008 said:
Thanks again, InteGrand, you're helping me so much! Really appreciate it! Just got a few more questions, if you don't mind. Also outta curiosity, how do you type those mathematical symbols on BoS? AND HOW R U SUCH A GOD AT MATHS? I mean, WHAT DO YOU DO?

Here are some hints.

$\noindent 6) (a) We can use calculus to find the stationary point, and sketch the individual graphs of $\ln x$ and $x$ to determine the sign of $\frac{\ln x}{x}$ at differing $x$-values. We can use the behaviour of $\ln x$ and $x$ near $x = 0$ to check whether there's an asymptote there. Also recall that polynomial so dominate logs, so the limit as $x\to \infty$ will be 0.$

$\noindent (b) Use the turning point.$

$\noindent 7. This can be done by repeated differentiation, it's like a HSC 4U Q.$

$\noindent 8. For the ones where the base involves $x$, you can convert it to a base of e to so that there is only a variable in the exponent, and then analyse the behaviour. For the fourth limit, the answer is just $\mathrm{e}^a$, and for the last two, recall that exponentials dominate polynomial so.$

$\noindent 9. (a) All you need to do is show $\cosh^2 t -\sinh^2 t=1$. Just use the definitions of these functions in terms of exponentials and the rest is just algebra.$

$\noindent (b) What two areas is the shaded area the difference of?$

$\noindent (c) Differentiate the formula in (b) and try and show it equals $\frac{1}{2}$, which will imply that $A(t) =\frac{t}{2}+C$ for some constant $C$. Then use a nice value of $t$ to show $C=0$.$

1008 · Mar 17, 2016

InteGrand said:
Here are some hints.

$\noindent 6) (a) We can use calculus to find the stationary point, and sketch the individual graphs of $\ln x$ and $x$ to determine the sign of $\frac{\ln x}{x}$ at differing $x$-values. We can use the behaviour of $\ln x$ and $x$ near $x = 0$ to check whether there's an asymptote there. Also recall that polynomial so dominate logs, so the limit as $x\to \infty$ will be 0.$

$\noindent (b) Use the turning point.$

$\noindent 7. This can be done by repeated differentiation, it's like a HSC 4U Q.$

$\noindent 8. For the ones where the base involves $x$, you can convert it to a base of e to so that there is only a variable in the exponent, and then analyse the behaviour. For the fourth limit, the answer is just $\mathrm{e}^a$, and for the last two, recall that exponentials dominate polynomial so.$

$\noindent 9. (a) All you need to do is show $\cosh^2 t -\sinh^2 t=1$. Just use the definitions of these functions in terms of exponentials and the rest is just algebra.$

$\noindent (b) What two areas is the shaded area the difference of?$

$\noindent (c) Differentiate the formula in (b) and try and show it equals $\frac{1}{2}$, which will imply that $A(t) =\frac{t}{2}+C$ for some constant $C$. Then use a nice value of $t$ to show $C=0$.$

Thanks for your reply InteGrand, the hints were really helpful. However, I still couldn't figure out 9c. Like I got to the point where A'(t) = 1/2, but I couldn't think of a value of t to show C = 0. How do I do that?

Also, I'd like your suggestion on these questions:

Just part b of this one below

InteGrand · Mar 17, 2016

1008 said:
Thanks for your reply InteGrand, the hints were really helpful. However, I still couldn't figure out 9c. Like I got to the point where A'(t) = 1/2, but I couldn't think of a value of t to show C = 0. How do I do that?

Also, I'd like your suggestion on these questions:

Just part b of this one below

To show C = 0 in Q 9(c), sub. t=0. Then the area becomes 0 since the point in the hyperbola is on the x-axis. Also, t/2 = 0 when t = 0. Hence C = 0.

InteGrand · Mar 17, 2016

1008 said:
Thanks for your reply InteGrand, the hints were really helpful. However, I still couldn't figure out 9c. Like I got to the point where A'(t) = 1/2, but I couldn't think of a value of t to show C = 0. How do I do that?

Also, I'd like your suggestion on these questions:

Just part b of this one below

54. Yes. By definition of the limit (using delta = epsilon/2), the limit is 3.

55. The limit is + infinity, by definition of right limits equalling +infinity (use delta = K/(K – 2)).

58. (b) This is just a numerical version of 58 (a). If you found the equilibrium concentration in (a) in terms of the parameters f, E and V, you can just sub. in the given values to find that for (b). To find when the concentration is within 0.01 of the equilibrium A, you just need to find for which t is |x(t) – A| = 0.01 (like in that question earlier with a terminal velocity).

1008 · Mar 19, 2016

InteGrand said:
54. Yes. By definition of the limit (using delta = epsilon/2), the limit is 3.

55. The limit is + infinity, by definition of right limits equalling +infinity (use delta = K/(K – 2)).

58. (b) This is just a numerical version of 58 (a). If you found the equilibrium concentration in (a) in terms of the parameters f, E and V, you can just sub. in the given values to find that for (b). To find when the concentration is within 0.01 of the equilibrium A, you just need to find for which t is |x(t) – A| = 0.01 (like in that question earlier with a terminal velocity).

Thanks so much InteGrand. Just another question, for one sided limits, when we're asked to find lim(x->a+) f(x) or lim(x-->a-) f(x), are we meant to just calculate f(a) in both cases if it is defined, or are we meant to find f(a) and f(-a) respectively?

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