Hey guys,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 =)
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 guys,Hey,
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
I've attached a write-out of the question and my answer.I'm taking a more closer look now and there's a few things I should comment on from my first glance:
~Declan Bolster
I'm taking a more closer look now and there's a few things I should comment on from my first glance:Thanks for the help. If you're not too busy simplifying it would be great!
- 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!Sure. :)
~Declan.B
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.
\[ \begin{align*} \tan(23^{\circ})&=\frac{a}{b} \tag{1} \\ \tan(20^{\circ})&=\frac{a+100}{b+c} \tag{2}\\
c^2+100^2&=1000^2 \tag{3} \end{align*} \]
Hi!
I'm stuck on a question about Bernoulli sequences and the binomial distribution from Unit 4 Topic 3 Methods.
If it says 'at least' does this mean P(X>9)? And could you calculate that by doing P(X>9)=1-P(0≤X≤9)? I tried that and I didn't achieve the answer in the textbook.
Thank you!
Hi!At first glance, I actually think that it is really simple. Technically speaking, "the distance from the foot of the cliff to a boat", and "the height of the cliff above sea level", are exactly the ingredients of a right-angled triangle formulation!
I have a question I'm stuck on from Unit 4 Topic 2 Methods. It's either really simple and I'm missing something or its generally a bit tricky. First time I read the question I thought of a right triangle but you can't assume that can you? Plus this Unit 4 methods, not Unit 1. ;D
So if you can't assume it is a right triangle then how do you find the height? It reminded me of the ambiguous case but I'm not sure how/if to use that property to answer the question. I would greatly appreciate the help as I am doing a revision of Unit 4 Topic 1&2 in class this week.
Thank you!
Hi!
I am a little confused as to what I need to do to complete this question- as it is a proof the textbook doesn't have a solution.
Thanks in advance :)
Hi!
Welcome to the forums!
It's hard to describe much of how you approach this without giving it away (which I will happily do if you're still confused but I think it's more rewarding and you learn more from help than answers). That being said:
- try starting by writing out the sine rule
- then look at c/sinC and either b/sinB or a/sinA (i.e. work with only 2 fractions at a time)
- Rearrange it so that you get [something]/c = [something]/sinC
- Do again but for whichever of b/sinB and a/sinA you didn't use
- Then combine the two equations you get together for the result you need.
Once you have part a part b should be pretty straightforward.
Hope this helps and please let me know if this clarifies things enough :)
Ahhhh thank you!!!
Hi!
Welcome to the forums!
It's hard to describe much of how you approach this without giving it away (which I will happily do if you're still confused but I think it's more rewarding and you learn more from help than answers). That being said:
- try starting by writing out the sine rule
- then look at c/sinC and either b/sinB or a/sinA (i.e. work with only 2 fractions at a time)
- Rearrange it so that you get [something]/c = [something]/sinC
- Do again but for whichever of b/sinB and a/sinA you didn't use
- Then combine the two equations you get together for the result you need.
Once you have part a part b should be pretty straightforward.
Hope this helps and please let me know if this clarifies things enough :)
whats the process behind solving these? Never seen them before i'm not sure where to start
P is a movable point on the line y=7-x
Find the coordinates of P when it is closest to the origin (0,0).
I have just finished year 10 and got a very unhappy mark for my methods exam (D+). I feel like it is due to me switching from general midway through the year and not being able to keep up with the pace. I want to get ahead on the topics for year 11 so that I am very confident in what I do. However I do not understand what I need to specifically revise on as the unit outlines only consist of hard to understand learning goals.
Does anyone know what specific topics and subtopics I have to revise for unit 1 at least? (I have attached the unit outlines given by my school)
Happy Holidays!
If you're given the rate of change for the trunk's growth, you can find out the amount it has grown by choosing an appropriate upper and lower bound then integrating the rate of change between those bounds.
Once you get this answer, all you really have to do is make sure you obtain a value for the mass and the end of the second stage that seems accurate (ie. by adding the result of the integral to the trunk radius then calculating the new mass.) Note that the density is irrelevant as the volume is almost analogous to the mass. Just remember to have your ratio the right way around as well.
Answer should be 4:1.
Welcome to the forums! ;DSorry for the long reply, but thanks so much for the advice!!
I can't really give specific tips since I don't have any details, but some general tips:
- Do your best to calm down; it may be difficult to do so (and it's especially easy for me to say) but maintaining a level head is going to do you so much more good. Also, this is your first exam, so don't feel the need to put so much pressure on yourself; base your future performance and tactics on what happens this time
- Remember that whoever is issuing the exam cannot (at least, in theory they shouldn't) test you on something you haven't learned. If there's something that doesn't seem quite right, break it down into smaller bits so you can use the knowledge you do have to solve the question
- Use any reading time you do get effectively
- Any other previous exam tips like if you get stuck move on, etc. also apply here.
Hope this helps!
Howdy.
Could someone please give me some tips on how to do well in a PSMT?
Many thanks. :)
Hi, my math methods exam is tomorrow and I was wondering if anyone knew how to solve this :)
Hi, I'm struggling with this question I'm not sure how the textbook gets to the answer. Would anyone be able to help?
Q: The Apache Orchard grows a very juicy apple called the Fuji apple. Fuji apples are picked and then sorted by diameter into three categories:
• small — diameter less than 60 mm
• jumbo — the largest 15% of the apples
• standard — all other apples.
f) Some apples are selected before sorting and are packed into bags of 6 to be sold at the front gate of the orchard. Determine the probability, correct to 4 decimal places, that one of these bags contains at least 2 jumbo apples.
The answer given is
I have a question with binomial probabilities and inputting into the calculator.
It is a binomial probability
with p = 0.18
and sample of 10
calculate Pr(P' < 0.2)
I have
P' = X/n = X/10
Pr(P' < 0.2) = Pr(X/10 < 0.2) = Pr(X<2)
The answers tell me in the calculator use binomial distribution function with n = 10, p=0.18
and the answer given is 0.4392.
But I can't get that I have no idea what to put in the x value part of the calculator, and should I be using binompdf or binom cdf?