Mathematics Question Thread

[RuiAce]

Thanks so much Rui!
But the answers say that it is 72.5u^2. Would hat then be wrong?

Well, taking a look at the diagram and the computations, 72.5u^2 would just be the combined area of just \( \triangle AMB\) and \(\triangle CMD\), i.e. the two triangles originally drawn and excluding the ones you have in pencil. However I feel as though the wording of the question is ambiguous then because these "self-intersecting quadrilaterals" are not a part of the course, so I interpreted the quadrilateral to actually include the new triangles you've drawn in.

So I blame the exam writers on this one. In the HSC, this kind of ambiguity would not occur.

[emmajb37]

Well, taking a look at the diagram and the computations, 72.5u^2 would just be the combined area of just \( \triangle AMB\) and \(\triangle CMD\), i.e. the two triangles originally drawn and excluding the ones you have in pencil. However I feel as though the wording of the question is ambiguous then because these "self-intersecting quadrilaterals" are not a part of the course, so I interpreted the quadrilateral to actually include the new triangles you've drawn in.

So I blame the exam writers on this one. In the HSC, this kind of ambiguity would not occur.

Awesome thanks again so much!!

[shaynec19]

Could someone please help me with the 1997 2u HSC exam, Question 10. b. iv.
I have found the x-coordinate, how do I find the y-coordinate.

[RuiAce]

Could someone please help me with the 1997 2u HSC exam, Question 10. b. iv.
I have found the x-coordinate, how do I find the y-coordinate.

At a quick glance, you could probably use the fact that \(P\) lies on the circle \(x^2+y^2=1\)?

[shaynec19]

At a quick glance, you could probably use the fact that \(P\) lies on the circle \(x^2+y^2=1\)?

Hi Rui,
Thankyou! I will try again, I already tried that and got nowhere but figured that because it was so long and messy, there must have been a better way...

[RuiAce]

Hi Rui,
Thankyou! I will try again, I already tried that and got nowhere but figured that because it was so long and messy, there must have been a better way...

Well the computations can be made cleaner if we're super clever about it

By doing this, we've completely avoided the quadratic formula
\[ \text{Now using }x^2+y^2=1\text{ we have}\\ \begin{align*}y^2 &= 1- x^2\\ &= 1 - \left(\frac{1-m^2}{1+m^2}\right)^2\\ &= \frac{(1+m^2)^2 - (1-m^2)^2}{(1+m^2)^2}\\ &= \frac{4m^2}{(1+m^2)^2}\\ \therefore y &= \frac{2m}{1+m^2} \end{align*} \]

But considering this is a pre-2001 paper, it's kinda expected that the question difficulty is notably higher.

[Rabi]

Hey I really need help with this question plz plz help.
"A water trough is 200 cm long and has a cross section of a right angled isosceles triangle. Show that when the depth of the water is x cm, the volume of water in the tank is 200x^2 cm^3. Water is poured in at a constant rate of 5 litres per minute. Find the rate at which the water level is rising when the depth is 30cm.

[RuiAce]

Hey I really need help with this question plz plz help.
"A water trough is 200 cm long and has a cross section of a right angled isosceles triangle. Show that when the depth of the water is x cm, the volume of water in the tank is 200x^2 cm^3. Water is poured in at a constant rate of 5 litres per minute. Find the rate at which the water level is rising when the depth is 30cm.

This is an MX1 related rates question. Please consult the MX1 thread if you require help with these concepts.

Having said that, this question was asked recently. You should also check the hints that were recently provided on addressing this question before indicating more of the problem.

[Georgakopoulou]

Hi! I am having trouble solving the attached question regarding the geometrical applications of calculus, could you please solve it for me? 

[RuiAce]

Hi! I am having trouble solving the attached question regarding the geometrical applications of calculus, could you please solve it for me? 

\[ \text{Let }C\text{ be the total cost of the trip. Then}\\ C = ct\\ \text{where }t\text{ is the time taken for the trip.} \]
\[ \text{Using the speed-distance-time formula we have }v = \frac{500}{t}\\ \text{so therefore }\boxed{C = \left( 150+\frac{v^2}{80} \right) \frac{500}{v}} \]
\begin{align*} C &=\frac{25}{4} \left( \frac{12000}{v}+v\right) \\ \frac{dC}{dv} &= \frac{25}{4} \left( -\frac{12000}{v^2}+1 \right)\end{align*}
You should be able to continue from here following the usual process. The minimum occurs at \(v = 20\sqrt{30} \) which is approximately 110.

[Georgakopoulou]

\[ \text{Let }C\text{ be the total cost of the trip. Then}\\ C = ct\\ \text{where }t\text{ is the time taken for the trip.} \]
\[ \text{Using the speed-distance-time formula we have }v = \frac{500}{t}\\ \text{so therefore }\boxed{C = \left( 150+\frac{v^2}{80} \right) \frac{500}{v}} \]
\begin{align*} C &=\frac{25}{4} \left( \frac{12000}{v}+v\right) \\ \frac{dC}{dv} &= \frac{25}{4} \left( -\frac{12000}{v^2}+1 \right)\end{align*}
You should be able to continue from here following the usual process. The minimum occurs at \(v = 20\sqrt{30} \) which is approximately 110.

Thank you so much!!!!!!!

[alexnero7]

Hi everyone,

Would someone be able to help me with a question? I am having trouble solving the attached question regarding the geometrical applications of calculus.

Thanks

[meerae]

Hi everyone,

Would someone be able to help me with a question? I am having trouble solving the attached question regarding the geometrical applications of calculus.

Thanks

Hi!
So what we know directly from the question is that y’=0 when x= -1 or 2
So all you would need to do is differentiate and do simultaneous equations.

Hope this helped!
meerae

[alexnero7]

Could someone please work this out?
Thanks

[meerae]

Could someone please work this out?
Thanks

Hey alexnero7!
I've attached the working out below, if you need any clarification please let me know as I tend to skip steps without even realising.
P.S for some reason its showing dark on my laptop - it should still be readable, if not please let me know!
Hope this helped!