Mathematics Question Thread

[MB_]

Yep Sure,
I have differentiated lnx to get 1/x. This is my gradient right?
I then used the point gradient formula with the x&y values.

Once you differentiate, you have to sub in the x value of the point specified to find the gradient

[shaynec19]

Oooohhhhh nooo...
What is wrong with me....
#holidaymaths
Thankyou very much!

[emilyyyyyyy]

Can someone please help me with what values or whatever i put into this trapezoidal rule formula (attached)? i've watched yt vids and looked in multiple textbooks but i don't really get how i'm meant to use this formula. Thanks!

[fun_jirachi]

Can someone please help me with what values or whatever i put into this trapezoidal rule formula (attached)? i've watched yt vids and looked in multiple textbooks but i don't really get how i'm meant to use this formula. Thanks!

If you look at the formula, you see that the first and terms are only multiplied by one while every other term is multiplied by two. Think about what the trapezoidal rule actually does over multiple applications; with two trapeziums side by side ie for the integral

when applying the trapezoidal rule with two applications you use

Notice that since the 'gaps' between the numbers are the same you can just reduce that to h/2 in the formula, where h is the distance between applications. Another thing to notice is that you count the first term (ie. y₀, or f(x) of the lower bound) and last term (ie. y_n, or f(x) of the upper bound) only once as they intuitively are on the 'outside' of the many trapeziums (ie. they are not included by two trapeziums). On the other hand, anything in between is counted twice because the function value is used twice by two different trapeziums. hence the 2(y₁+y₂+y₃+...+y_n-1).

Basically for n applications of the trapezoidal rule over the curve f(x), subtract the lower bound from the upper bound, divide n into that number to get another number, say h. h+the lower bound becomes your next point of reference, 2h+lowerbound is the next and so on until you get to the upper bound. Find f(x) of these values to get what you need to sub in. f(x) of the lower bound and upper bound will get counted once only (first and last) and everything else that you've figured out will get counted twice. h is also your height that you sub into that formula.

If you need a better/quick way to memorise it, I remembered it like this; height/2 x (first + last + 2(everything else)) which is the same thing as what you've given, but much easier to remember.

Hope this helps! ")

[shaynec19]

Hi,
Can someone please explain why when differentiating this exponential, that ln2 becomes part of the coefficient? I thought it would just be the 3, as the 3x-4 is the term with the pronumeral.


Thank You for anyone's help.

[RuiAce]

Hi,
Can someone please explain why when differentiating this exponential, that ln2 becomes part of the coefficient? I thought it would just be the 3, as the 3x-4 is the term with the pronumeral.


Thank You for anyone's help.

\(\ln 2\) is a constant.

That last step was redundant but I only did it to illustrate that what comes next is we expand the \(\ln 2\) term into the bracket.
\[ e^{\ln 2 \times (3x-4)} = e^{3x\ln 2 - 4\ln 2} \]
Therefore the coefficient on \(x\) is \(3\ln 2\), justifying why that's what comes down after differentiating

[shaynec19]

\(\ln 2\) is a constant.

That last step was redundant but I only did it to illustrate that what comes next is we expand the \(\ln 2\) term into the bracket.
\[ e^{\ln 2 \times (3x-4)} = e^{3x\ln 2 - 4\ln 2} \]
Therefore the coefficient on \(x\) is \(3\ln 2\), justifying why that's what comes down after differentiating

Aaah Thanks Rui!
Didn't think about expanding the brackets! Makes perfect sense now.
Cheers
Shayne

[shaynec19]

Could Someone please help me with finding the fraction that leads to finding primitive function of:

Is there any tips and and tricks to finding it? I know its 3/2 in this case, and I can understand why it's there but can't seem to be able to reverse engineer it.

[RuiAce]

Could Someone please help me with finding the fraction that leads to finding primitive function of:

Is there any tips and and tricks to finding it? I know its 3/2 in this case, and I can understand why it's there but can't seem to be able to reverse engineer it.

\[ \text{At the end of the day, somehow you're trying to invoke that}\\ \int \frac{f^\prime(x)}{f(x)}\,dx = \ln [f(x)]+C.\\ \text{It's your job to actually make }f^\prime(x)\text{ appear.} \]

\[ \text{That manipulation in line 1 where we force }2x\text{ to appear}\\ \text{is the crucial step.} \]

[shaynec19]

Could someone please help with this question;

[RuiAce]

Could someone please help with this question;

\begin{align*}y&= \log_e x\\ x&= e^y\\ x^2 &= e^{2y} \end{align*}
So your volume integral will be \( \pi\int_1^3 e^{2y}\,dy \).

[shaynec19]

\begin{align*}y&= \log_e x\\ x&= e^y\\ x^2 &= e^{2y} \end{align*}
So your volume integral will be \( \pi\int_1^3 e^{2y}\,dy \).

Thanks Rui!
Also, thank you for your advice with the exam!

[emmajb37]

Hi, I am struggling with this past trial question please help! Western Region Trial Exam 2005 Question 5 a) ii)
I have attached a photo.
Why does geometry exist 

[RuiAce]

Thanks Rui!
Also, thank you for your advice with the exam!

Haha no worries. Your exam was certainly a weird case. I feel a bit bad for you given the circumstances around it but there was still a very subtle mistake anyhow

Hi, I am struggling with this past trial question please help! Western Region Trial Exam 2005 Question 5 a) ii)
I have attached a photo.
Why does geometry exist 

Well there's a bit of trigonometry involved on top of the geometry here.
\[ \text{Part i) should be obvious, given that }AB \parallel CD\\ \text{so clearly the two alternate angles will help complete the equiangular test.} \]
\[ \text{Now as a consequence of the similar triangles, we have the proportional sides}\\ \frac{AM}{DM} = \frac{MB}{MC} = \frac{AB}{DC}.\\ \text{But we're also told that }\frac{AB}{CD} = \frac25.\\ \text{This consequently implies that }\frac{AM}{DM} = \frac25 \implies \boxed{DM = \frac52 AM}\\ \text{and also implies that }\frac{MB}{MC} = \frac25 \implies \boxed{CM = \frac52 BM}.\]
\[ \text{To invoke areas, let }\angle AMB = \theta.\text{ Then from vertically opposite angles, }\angle CMD = \theta\\ \text{and from angles on straight angles, }\angle AMC = \angle BMD = 180^\circ - \theta.\]
\[ \text{Consequently }\sin \angle AMB = \sin \angle CMD = \sin \theta\\ \text{and }\sin \angle AMC = \sin \angle BMD = \sin (180^\circ - \theta).\\ \text{However using our ASTC identities, we know that }\boxed{\sin(180^\circ - \theta) = \sin \theta}\\ \text{so very conveniently, we have}\\ \boxed{\sin \angle AMB = \sin \angle AMC = \sin \angle BMD = \sin \angle CMD = \sin \theta} \]
\[ \text{This set-up was used because since}\operatorname{Area}_{\triangle AMB} = 10,\\ \text{we have }\boxed{\frac12 \, AM \, BM\sin \theta = 10 }.\\ \text{Note that the area we require can be thought of as}\\ \boxed{\operatorname{Area}_{ABCD} = \operatorname{Area}_{\triangle AMB} + \operatorname{Area}_{\triangle AMC} + \operatorname{Area}_{\triangle BMD} + \operatorname{Area}_{\triangle CMD}}\\ \text{and we now have the ingredients to find the areas of all four triangles.} \]
\begin{align*} \operatorname{Area}_{\triangle AMB} &= \frac12 \, AM\, BM \sin \theta = 10\\ \operatorname{Area}_{\triangle AMC} &= \frac12\, AM\, CM\sin\theta = \frac12\,AM \left(\frac52BM\right)\sin\theta = 25\\ \operatorname{Area}_{\triangle BMD} &= \frac12\, BM\, DM\sin\theta = \frac12\,BM \left(\frac52AM\right)\sin\theta = 25\\ \operatorname{Area}_{\triangle CMD} &= \frac12\, CM\, DM\sin\theta = \frac12\left( \frac52 AM\right) \left(\frac52BM\right)\sin\theta =62.5 \end{align*}
The final answer is hence just the sum of these, which is 122.5 units².

[emmajb37]

Haha no worries. Your exam was certainly a weird case. I feel a bit bad for you given the circumstances around it but there was still a very subtle mistake anyhowWell there's a bit of trigonometry involved on top of the geometry here.
\[ \text{Part i) should be obvious, given that }AB \parallel CD\\ \text{so clearly the two alternate angles will help complete the equiangular test.} \]
\[ \text{Now as a consequence of the similar triangles, we have the proportional sides}\\ \frac{AM}{DM} = \frac{MB}{MC} = \frac{AB}{DC}.\\ \text{But we're also told that }\frac{AB}{CD} = \frac25.\\ \text{This consequently implies that }\frac{AM}{DM} = \frac25 \implies \boxed{DM = \frac52 AM}\\ \text{and also implies that }\frac{MB}{MC} = \frac25 \implies \boxed{CM = \frac52 BM}.\]
\[ \text{To invoke areas, let }\angle AMB = \theta.\text{ Then from vertically opposite angles, }\angle CMD = \theta\\ \text{and from angles on straight angles, }\angle AMC = \angle BMD = 180^\circ - \theta.\]
\[ \text{Consequently }\sin \angle AMB = \sin \angle CMD = \sin \theta\\ \text{and }\sin \angle AMC = \sin \angle BMD = \sin (180^\circ - \theta).\\ \text{However using our ASTC identities, we know that }\boxed{\sin(180^\circ - \theta) = \sin \theta}\\ \text{so very conveniently, we have}\\ \boxed{\sin \angle AMB = \sin \angle AMC = \sin \angle BMD = \sin \angle CMD = \sin \theta} \]
\[ \text{This set-up was used because since}\operatorname{Area}_{\triangle AMB} = 10,\\ \text{we have }\boxed{\frac12 \, AM \, BM\sin \theta = 10 }.\\ \text{Note that the area we require can be thought of as}\\ \boxed{\operatorname{Area}_{ABCD} = \operatorname{Area}_{\triangle AMB} + \operatorname{Area}_{\triangle AMC} + \operatorname{Area}_{\triangle BMD} + \operatorname{Area}_{\triangle CMD}}\\ \text{and we now have the ingredients to find the areas of all four triangles.} \]
\begin{align*} \operatorname{Area}_{\triangle AMB} &= \frac12 \, AM\, BM \sin \theta = 10\\ \operatorname{Area}_{\triangle AMC} &= \frac12\, AM\, CM\sin\theta = \frac12\,AM \left(\frac52BM\right)\sin\theta = 25\\ \operatorname{Area}_{\triangle BMD} &= \frac12\, BM\, DM\sin\theta = \frac12\,BM \left(\frac52AM\right)\sin\theta = 25\\ \operatorname{Area}_{\triangle CMD} &= \frac12\, CM\, DM\sin\theta = \frac12\left( \frac52 AM\right) \left(\frac52BM\right)\sin\theta =62.5 \end{align*}
The final answer is hence just the sum of these, which is 122.5 units².

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