Help with Applied calculus needed before tmz

Doylers

Hi Im trying to pass a third level calculus paper tomorrow and I need some help on how to do the problems so if anyone solve the problems below showing there work I will forever in your debt. Im looking for help with Q5 (a), Q1 (a) and Q2 (A) and (b). Thanks alot

rainbow kirby

What have you done so far for each of those problems?

sponsoredwalk

Well for Question 5, on Simpsons Rule, you've been given the formula but
I bet you don't know how to apply it, here is a nice small tutorial.

Question 1 (a) is just a standard calculus question, if you don't understand
what is being asked in it I'd doubt you understand part (b) either.
You'd have to show some attempt to let us know what your problems are...

Question 2 involves some thought about functions, again, some input on
what you've done would be helpful...

Doylers

Ok im attaching Q1 (a) and (b) with this post.

For Q2 (a) how would I expand X^3 out so I can simplify the equation?
Seriously thanks for any help at all

Doylers

Attached Q2(B) part 1

rainbow kirby

You've got the right idea with Q1(a), but it's not quite right.

[latex]
V(t) = 2 \ln{t}
[/latex]
[latex]
\frac{dV}{dt} = 2 \times \frac{1}{t}
[/latex]
You've got this much right anyway.

The mistake you've made is that you've substituted [latex]t = 2[/latex] instead of [latex]t = 1[/latex], and also that you've missed out on the fact that it's 2 multiplied by [latex]\frac{1}{t}[/latex] and not [latex]2\frac{1}{t}[/latex].

Doylers

rainbow kirby wrote: »

You've got the right idea with Q1(a), but it's not quite right.

[latex]
V(t) = 2 \ln{t}
[/latex]
[latex]
\frac{dV}{dt} = 2 \times \frac{1}{t}
[/latex]
You've got this much right anyway.

The mistake you've made is that you've substituted [latex]t = 2[/latex] instead of [latex]t = 1[/latex], and also that you've missed out on the fact that it's 2 multiplied by [latex]\frac{1}{t}[/latex] and not [latex]2\frac{1}{t}[/latex].

Ok I see what your saying there that makes sense.

so when i multiply it out does it just become 2 over 1?

rainbow kirby

Yes it does.

Doylers

Ok cool, how would i go about solving: Find the equation of the tangent at t: = l.Show your answer on a graph. The only way I know of is using the formula y-y1=m(x-x1) but I only have one point and no slope.

sponsoredwalk

Your answer is wrong, you plugged in t = 2 when the question asked you to
plug in t = 1 , also part (ii) of the question asked you to find the
equation of the tangent line at this moment in time & to draw it.
[latex] y - y_0 \ = \ f'(x_0)(x - x_0) [/latex] is the general idea, you knew that right?
Btw, this equation is also extremely helpful in the question on the
paper that asks you to derive Newton's Method

Question (b) falls out of basic kinematics of constant acceleration,
if you have studied this out of a calculus book I feel sorry for you
I hadn't a clue what this was talking about in a calculus book but it falls
out of basic physics easy, you want to find out when x = 0 after
integrating the acceleration function twice to get the position function.
BAsically, take the function they've given you and integrate it twice, you'll
have to be careful because the constants you get in both integrations
matter normally but really you can discard the constant in the
velocity function because your initial velocity is zero as the question indicates!

Part (c) of Q1 is just a position function, take 2 derivatives to get the
acceleration & follow the instructions of the question (you knew all this already right?).

Question 2 involves a few subtle techniques, you don't understand how to
get vertical and horizontal asymptotes at all do you?

This is a lot to take in, I really hope you aren't new to this & aren't taking
a serious exam tomorrow!!! Let us know quick & we'll see what we can do!

neilk32

Hey Sean nice thread :P Will help me too

rainbow kirby

Doylers wrote: »

Ok cool, how would i go about solving: Find the equation of the tangent at t: = l.Show your answer on a graph. The only way I know of is using the formula y-y1=m(x-x1) but I only have one point and no slope.

You do have a slope - you just got it in the first part of the question!

Doylers

sponsoredwalk wrote: »

Your answer is wrong, you plugged in t = 2 when the question asked you to
plug in t = 1 , also part (ii) of the question asked you to find the
equation of the tangent line at this moment in time & to draw it.
[latex] y - y_0 \ = \ f'(x_0)(x - x_0) [/latex] is the general idea, you knew that right?
Btw, this equation is also extremely helpful in the question on the
paper that asks you to derive Newton's Method

Question (b) falls out of basic kinematics of constant acceleration,
if you have studied this out of a calculus book I feel sorry for you
I hadn't a clue what this was talking about in a calculus book but it falls
out of basic physics easy, you want to find out when x = 0 after
integrating the acceleration function twice to get the position function.
BAsically, take the function they've given you and integrate it twice, you'll
have to be careful because the constants you get in both integrations
matter normally but really you can discard the constant in the
velocity function because your initial velocity is zero as the question indicates!

Part (c) of Q1 is just a position function, take 2 derivatives to get the
acceleration & follow the instructions of the question (you knew all this already right?).

Question 2 involves a few subtle techniques, you don't understand how to
get vertical and horizontal asymptotes at all do you?

This is a lot to take in, I really hope you aren't new to this & aren't taking
a serious exam tomorrow!!! Let us know quick & we'll see what we can do!

Thanks for the reply, yup a repeat exam tomorrow at 9 lol I never had a book we went to class every day and she said write these notes off the board and I would write 20 odd pages a day without even knowing what I was writing because she never explained things. The formula you posted what would be put into it from the question?

Q2 no I have no idea about them but if you can show me what to do ill pick it up fast and as for part b of question 2 I have posted the first part to see if im on the right track.

Doylers

rainbow kirby wrote: »

You do have a slope - you just got it in the first part of the question!

Oh ok since they touch at t=1 they have the same slope? To get a second point do I put t=1 into the original equation?

rainbow kirby

You can get a point by using [latex]t = 1[/latex] to get [latex]V = 2 \ln (1) = 0[/latex].

Use slope [latex]m = 2[/latex] and point [latex](1,0)[/latex] in the formula for the equation of a line and that will give you the equation of the tangent.

sponsoredwalk

Wow! Sounds like 6 years of secondary school for me lol!
My advice to you is to get away from the teacher & teach yourself!
Start Tonight!!!

www.khanacademy.org

Watch as many videos as you can tonight in the remaining time,
& get up EARLY to review - I mean it - the videos are at the level of
the questions in the pdf you put up & will give you the intuition behind what
you're doing.

There is a video on finding the equation to the tangent line here

In the video he writes that he's finding the point (1,e) well you're looking
for the point (1,0) because when you plug 1 into your function you
get the answer of 0 for the y value, remember ln(1) = 0

y - 0 = f'(0)(x - 0) = watch the video to understand it then look at the other videos to get an idea

Doylers

sponsoredwalk wrote: »

Wow! Sounds like 6 years of secondary school for me lol!
My advice to you is to get away from the teacher & teach yourself!
Start Tonight!!!

www.khanacademy.org

Watch as many videos as you can tonight in the remaining time,
& get up EARLY to review - I mean it - the videos are at the level of
the questions in the pdf you put up & will give you the intuition behind what
you're doing.

There is a video on finding the equation to the tangent line here

In the video he writes that he's finding the point (1,e) well you're looking
for the point (1,0) because when you plug 1 into your function you
get the answer of 0 for the y value, remember ln(1) = 0

y - 0 = f'(0)(x - 0) = watch the video to understand it then look at the other videos to get an idea

Ok i watched his video an I get what your all saying. My final qustion about Q1 is the formula people are using is y - 0 = f'(0)(x - 0), and in the video he uses y=(2e)x+b , im used to the formula y-y1=m(x-x1) why isnt that being used? Can I use it or will I learn off the other one?

sponsoredwalk

He explained what he was doing, it's the exact same meaning.

y - y_1 = m(x - x_1) can be written as

y = mx - mx_1 + y_1

and if we call

b = y_1 - mx_1

we can rewrite it as y = mx + b

He even does this in the video to show you, but don't worry.

The slope m is just the derivative of a function at that point.

Your function f(t) = 2ln(t) has a derivative of 2/t

If you want to find the equation of the tangent line at t = 1 use the
equation y - y_1 = m(x - x_1)

This is the same thing as f(x) - f(x_1) = f ' (x_1) (x - x_1)

Your question actually uses the letter t so we can rewrite it all as

equation f(t) - f(t_1) = m(t - t_1)


t & f(t) are variable, we want f(t_1) and t_1, well t_1 = 1 as the question asked us to pick, so if we want f(t_1) we just plug 1 into 2ln(t) and we
get 0.

f(t) - 0 = m(t - 0)

You found m = f ' (t) = 2/t so if t = 1 then f ' (t) = 2

f(t) - 0 = 2(t - 0)

f(t) = 2t

that is the equation of the tangent line at t = 1.

You did the right thing for (b), I told you how to do c.

You'll have to watch a load of videos to understand question 2,
here is a graph of it, it might
help you when you've watched a few videos
question 5 I've given you a tutorial how to answer it.

Good luck tomorrow!