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January 22, 2022, 06:57:26 am AuthorTopic: QCE Maths Methods Questions Thread  (Read 14446 times)

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Joseph41

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If you have general questions about the QCE Maths Methods course (both Units 1&2 and 3&4) or how to improve in certain areas, this is the place to ask! 👌

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Declan.B

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« Reply #1 on: November 03, 2019, 07:05:41 pm »
+1
Hey guys,
I'm just trying to differentiate the following function:
y=1000×b^t×e^t (c(a(1-b^t×e^t ))^d ).
The "t" values are the x values (I put them as t to prevent confusion with multiplication symbols)
I tried to break up sections of it and solve using product rule

product rule:let f(x)= 1000c(be)^t,let g(x)=a^d (1-(be)^dt)
y' =f^' (x)g(x)+f(x) g^' (x)
let z(x)=f(x)
let 1000cb^t=a(x),e^t=b(x)
a^' (x)=1000×c×ln⁡(b)×b^t
b^' (x)=e^t×ln⁡(e)=e^t

product rule: z^' (x)=a^' (x)×b(x)+a(x) b^' (x)
z^' (x)=(1000×c×ln⁡(b)×b^t ) e^t+(1000×c×b^t ) e^t
z(x)=f(x)∴z^' (x)=f^' (x)
f^' (x)=(1000×c×ln⁡(b)×b^t ) e^t+(1000×c×b^t ) e^t
f^' (x)=e^t×1000×c×b^t (ln⁡(b)+1)

let p(x)=g(x)
p(x)=a^d (1-(be)^dt )
p(x)=(a^d-a^d (be)^dt

p^' (x)=(da^(d-1)-da^(d-1)×ln⁡(b)×d(be)^(d-1×t) )
p^' (x)=da^(d-1) (1- ln⁡(b)×d(be)^(d-1×t) )
p(x)=g(x)∴p^' (x)=g^' (x)
g^' (x)=da^(d-1) (1- ln⁡(b)×d(be)^(d-1×t) )

y^'=f^' (x)g(x)+f(x) g^' (x)
y^'=[e^t×1000×c×b^t (ln⁡(b)+1)][a^d (1-(be)^dt )]+[1000c(be)^t ][da^(d-1) (1- ln⁡(b)×d(be)^(d-1×t) )]

Sorry if it's a bit messy (I'm new to differentiating exponentials) but could someone let me know if my working is correct. I didn't want to simplify my answer if it wasn't correct. Also, the additional coefficients are just variables that I wanted to do a general case for before substituting.
~Declan.B

DrDusk

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« Reply #2 on: November 03, 2019, 07:34:27 pm »
0
Hey guys,
I'm just trying to differentiate the following function:
y=1000×b^t×e^t (c(a(1-b^t×e^t ))^d ).
The "t" values are the x values (I put them as t to prevent confusion with multiplication symbols)
I tried to break up sections of it and solve using product rule

product rule:let f(x)= 1000c(be)^t,let g(x)=a^d (1-(be)^dt)
y' =f^' (x)g(x)+f(x) g^' (x)
let z(x)=f(x)
let 1000cb^t=a(x),e^t=b(x)
a^' (x)=1000×c×ln⁡(b)×b^t
b^' (x)=e^t×ln⁡(e)=e^t

product rule: z^' (x)=a^' (x)×b(x)+a(x) b^' (x)
z^' (x)=(1000×c×ln⁡(b)×b^t ) e^t+(1000×c×b^t ) e^t
z(x)=f(x)∴z^' (x)=f^' (x)
f^' (x)=(1000×c×ln⁡(b)×b^t ) e^t+(1000×c×b^t ) e^t
f^' (x)=e^t×1000×c×b^t (ln⁡(b)+1)

let p(x)=g(x)
p(x)=a^d (1-(be)^dt )
p(x)=(a^d-a^d (be)^dt

p^' (x)=(da^(d-1)-da^(d-1)×ln⁡(b)×d(be)^(d-1×t) )
p^' (x)=da^(d-1) (1- ln⁡(b)×d(be)^(d-1×t) )
p(x)=g(x)∴p^' (x)=g^' (x)
g^' (x)=da^(d-1) (1- ln⁡(b)×d(be)^(d-1×t) )

y^'=f^' (x)g(x)+f(x) g^' (x)
y^'=[e^t×1000×c×b^t (ln⁡(b)+1)][a^d (1-(be)^dt )]+[1000c(be)^t ][da^(d-1) (1- ln⁡(b)×d(be)^(d-1×t) )]

Sorry if it's a bit messy (I'm new to differentiating exponentials) but could someone let me know if my working is correct. I didn't want to simplify my answer if it wasn't correct. Also, the additional coefficients are just variables that I wanted to do a general case for before substituting.
~Declan.B
Sorry but I can't make sense of it written like this =( If you could perhaps write it down physically and upload it as an attachment or even better write it in LateX(any of which would be fine), then I could help you =)
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RuiAce

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« Reply #3 on: November 03, 2019, 08:11:50 pm »
0
Hey guys,
I'm just trying to differentiate the following function:
y=1000×b^t×e^t (c(a(1-b^t×e^t ))^d ).
The "t" values are the x values (I put them as t to prevent confusion with multiplication symbols)
I tried to break up sections of it and solve using product rule

product rule:let f(x)= 1000c(be)^t,let g(x)=a^d (1-(be)^dt)
y' =f^' (x)g(x)+f(x) g^' (x)
let z(x)=f(x)
let 1000cb^t=a(x),e^t=b(x)
a^' (x)=1000×c×ln⁡(b)×b^t
b^' (x)=e^t×ln⁡(e)=e^t

product rule: z^' (x)=a^' (x)×b(x)+a(x) b^' (x)
z^' (x)=(1000×c×ln⁡(b)×b^t ) e^t+(1000×c×b^t ) e^t
z(x)=f(x)∴z^' (x)=f^' (x)
f^' (x)=(1000×c×ln⁡(b)×b^t ) e^t+(1000×c×b^t ) e^t
f^' (x)=e^t×1000×c×b^t (ln⁡(b)+1)

let p(x)=g(x)
p(x)=a^d (1-(be)^dt )
p(x)=(a^d-a^d (be)^dt

p^' (x)=(da^(d-1)-da^(d-1)×ln⁡(b)×d(be)^(d-1×t) )
p^' (x)=da^(d-1) (1- ln⁡(b)×d(be)^(d-1×t) )
p(x)=g(x)∴p^' (x)=g^' (x)
g^' (x)=da^(d-1) (1- ln⁡(b)×d(be)^(d-1×t) )

y^'=f^' (x)g(x)+f(x) g^' (x)
y^'=[e^t×1000×c×b^t (ln⁡(b)+1)][a^d (1-(be)^dt )]+[1000c(be)^t ][da^(d-1) (1- ln⁡(b)×d(be)^(d-1×t) )]

Sorry if it's a bit messy (I'm new to differentiating exponentials) but could someone let me know if my working is correct. I didn't want to simplify my answer if it wasn't correct. Also, the additional coefficients are just variables that I wanted to do a general case for before substituting.
~Declan.B

Hey,

Before we jump into this, I just want to check if this was the function you were after?
$y = 100 b^t e^t \left( c\left(a (1-b^t e^t) \right)^d\right)$
That, and:
a) you're after a derivative in terms of $t$?
b) do I assume that $a$, $b$, $c$ and $d$ are constants?

(It may be really hard to type it out without LaTeX ngl. If you feel it's too hard, please feel welcome to just post pictures/screenshots instead. « Last Edit: November 03, 2019, 08:14:39 pm by RuiAce »  Declan.B

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« Reply #4 on: November 03, 2019, 08:51:50 pm »
+1
I've attached a write-out of the question and my answer.
~Declan Bolster

RuiAce

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« Reply #5 on: November 04, 2019, 01:01:30 pm »
+3
I've attached a write-out of the question and my answer.
~Declan Bolster
I'm taking a more closer look now and there's a few things I should comment on from my first glance:
- You're already using $a$ and $b$ for constants already. It's not recommended to now define stuff like $a(x)$ $b(x)$, where you use the same letter.
- Your derivative is in terms of $t$. Use functions in terms of $t$, not $x$.
- $1 - (be)^{dt}$ isn't the same thing as $(1-be^t)^d$.

So I feel the problem got made a bit more convoluted than what it had to be! This is how I would've done it. First rewrite:
$y = 1000 a^d c (be)^t \left( 1- (be)^t\right)^d.$
Since $a$, $c$ and $d$ are constants here, we can just leave the $1000 a^d c$ bit in front, and focus on everything else for the product rule.

Set up for the product rule:
\begin{align*}
f(t) &= (be)^t \\
g(t) &= \left( 1 - (be)^t\right)^d.
\end{align*}
Judging by your working out, I'm assuming you're allowed to use the rule $\boxed{\frac{d}{dx}a^x = a^x \ln (a)}$. Applying it here, we have
\begin{align*}
f^\prime (t) &= (be)^t \ln (be)\\
g^\prime(t) &= (be)^t\ln (be) \times d\left(1 - (be)^t \right)^{d-1}
\end{align*}
having used the chain rule on the second computation.

Note that $\ln(be) = \ln(b)+\ln(e) = \ln(b)+1$, which does not conflict with some of the stuff I see you did there.

Hence, putting everything together, the derivative is
\begin{align*}
\frac{dy}{dt} &= 1000 a^d c \left[g(t)f^\prime(t) + f(t)g^\prime(t) \right]\\
&= 1000 a^d c\left[ \left(1 - (be)^t\right)^d (be)^t \ln (be) + (be)^t (be)^t \ln (be) d \left(1-(be)^t \right)^{d-1} \right]
\end{align*}
This certainly can be simplified further, but it's still going to be ugly for sure. If you want me to simplify it, feel free to ask   Declan.B

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« Reply #6 on: November 04, 2019, 07:56:33 pm »
0
I'm taking a more closer look now and there's a few things I should comment on from my first glance:
- You're already using $a$ and $b$ for constants already. It's not recommended to now define stuff like $a(x)$ $b(x)$, where you use the same letter.
- Your derivative is in terms of $t$. Use functions in terms of $t$, not $x$.
- $1 - (be)^{dt}$ isn't the same thing as $(1-be^t)^d$.

So I feel the problem got made a bit more convoluted than what it had to be! This is how I would've done it. First rewrite:
$y = 1000 a^d c (be)^t \left( 1- (be)^t\right)^d.$
Since $a$, $c$ and $d$ are constants here, we can just leave the $1000 a^d c$ bit in front, and focus on everything else for the product rule.

Set up for the product rule:
\begin{align*}
f(t) &= (be)^t \\
g(t) &= \left( 1 - (be)^t\right)^d.
\end{align*}
Judging by your working out, I'm assuming you're allowed to use the rule $\boxed{\frac{d}{dx}a^x = a^x \ln (a)}$. Applying it here, we have
\begin{align*}
f^\prime (t) &= (be)^t \ln (be)\\
g^\prime(t) &= (be)^t\ln (be) \times d\left(1 - (be)^t \right)^{d-1}
\end{align*}
having used the chain rule on the second computation.

Note that $\ln(be) = \ln(b)+\ln(e) = \ln(b)+1$, which does not conflict with some of the stuff I see you did there.

Hence, putting everything together, the derivative is
\begin{align*}
\frac{dy}{dt} &= 1000 a^d c \left[g(t)f^\prime(t) + f(t)g^\prime(t) \right]\\
&= 1000 a^d c\left[ \left(1 - (be)^t\right)^d (be)^t \ln (be) + (be)^t (be)^t \ln (be) d \left(1-(be)^t \right)^{d-1} \right]
\end{align*}
This certainly can be simplified further, but it's still going to be ugly for sure. If you want me to simplify it, feel free to ask Thanks for the help. If you're not too busy simplifying it would be great!
~Declan.B

RuiAce

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« Reply #7 on: November 05, 2019, 02:19:12 pm »
+3
Thanks for the help. If you're not too busy simplifying it would be great!
~Declan.B
Sure. Basically the simplifying relies on some careful factorisation. Firstly it should be clear that $(be)^t$ and $\ln (be)$ are common factors, so we first pull them out.
\begin{align*}
\frac{dy}{dt} &= 1000a^d c (be)^t \ln (be) \left[ \left( 1 - (be)^t\right)^d + (be)^t d \left(1-(be)^t \right)^{d-1} \right]\\
&= 1000a^d c(be)^t \left( \ln(b) + 1 \right)\left[ \left( 1 - (be)^t\right)^d + (be)^t d \left(1-(be)^t \right)^{d-1} \right]
\end{align*}
A common thing with product rule computations involving powers is that a power can also be factored out. Here, observe that $\left(1-(be)^t\right)^{d-1}$ can be factored out.

For the second term, this should be clear. For the first term, the reason this works is because of index laws: $\left( 1-(be)^t\right)^d = \left( 1-(be)^t\right) \times \left( 1-(be)^t\right)^{d-1}$.
\begin{align*}
\frac{dy}{dt} &= 1000 a^d c (be)^t (\ln(b) +1) \left( 1-(be)^t\right)^{d-1} \left[1 + (be)^t d \right].
\end{align*}
Once here, I can't see any more simplification possible   A.Rose

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« Reply #8 on: March 12, 2020, 07:33:06 pm »
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Hello!
Would anyone be able to help me with this Unit 4 Topic 2 (Trig with sine and cosine rule) Question:

"A surveyor measures the angle of elevation of the top of a mountain from a point at sea level as 20∘. She then travels 1000 m along a road that slopes uniformly uphill towards the mountain. From this point, which is 100 m above sea level, she measures the angle of elevation as 23∘. Find the height of the mountain above sea level, correct to the nearest metre."

I mainly need to make sure I am drawing it correctly and then I can see how I go with finding the height.

Thank you!

fun_jirachi

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« Reply #9 on: March 13, 2020, 12:24:00 am »
+1
Hey there!

This should be a rough diagram Spoiler Hope this helps Spoiler
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A.Rose

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« Reply #10 on: March 13, 2020, 10:30:46 pm »
+1
Thank you yes that did help!
Although I got h = 1839 m plus 100 for height above sea level would be 1939 m and I don't know if that is right? Would you be able to check, please?

Also if possible there is another question I am stuck on:
From a clifftop 11 m above sea level, two boats are observed. One has an angle of depression of 45∘ and is due east, the other an angle of depression of 30∘ on a bearing of 120∘. Calculate the distance between the boats.
I'm a bit confused about where the 2nd boat is positioned - can you help me with the drawing?
Thank you!

fun_jirachi

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« Reply #11 on: March 13, 2020, 11:47:12 pm »
+2
That's odd, I've gotten about 410m - 1939 seems a bit big, considering the angles given are roughly equal and the height increase is only 100m This one is a little tougher to draw - it's a 3D diagram on a 2D plane! I don't trust my hand to draw things that other people can understand, so hopefully my MS Paint work is good enough Spoiler Hope this helps Spoiler
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Specialist_maths Re: QCE Maths Methods Questions Thread
« Reply #12 on: March 14, 2020, 09:01:46 am »
+3
Thank you yes that did help!
Although I got h = 1839 m plus 100 for height above sea level would be 1939 m and I don't know if that is right? Would you be able to check, please?

Hi A.Rose,

I also got 1939 m for the height of the mountain, by solving the following equations simultaneously.

\begin{align*} \tan(23^{\circ})&=\frac{a}{b} \tag{1} \\ \tan(20^{\circ})&=\frac{a+100}{b+c} \tag{2}\\ c^2+100^2&=1000^2 \tag{3} \end{align*}
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fun_jirachi

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« Reply #13 on: March 14, 2020, 09:22:32 am »
+1
Hi A.Rose,

I also got 1939 m for the height of the mountain, by solving the following equations simultaneously.

\begin{align*} \tan(23^{\circ})&=\frac{a}{b} \tag{1} \\ \tan(20^{\circ})&=\frac{a+100}{b+c} \tag{2}\\ c^2+100^2&=1000^2 \tag{3} \end{align*}

I've made an error - you would both be correct! Just realised I misread my working from one line to the next Spoiler
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fun_jirachi

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« Reply #14 on: March 19, 2020, 10:54:30 pm »
+3
Hey there!

I think the reason that you're a bit stuck on this question is that it's impossible to have a triangle with those specified dimensions, then have the third side equal to 325m! The question should really have angle ACB (which essentially means you have a single application of the cosine rule to do to get the answer), instead of angle CAB, which results in the third side being 677m.

Hope this helps Spoiler
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