Find the derivative and graphmatica

[Teggle]

I've been given the question of:

 

Derive the following function to find the velocity and acceleration: s=0.3cos(pi/4*t)

 

I am unsure of how to do this properly with this particuar function.

 

It would also help if someone could let me know how to put this function into Graphmatica. I know you have to go to Tools and then down to Functions to enter it into the program, but I don't know how to write it so it will work, as an error keeps coming up when I try to enter it as I have written above.

Thanks to all who can help.

[mooeypoo]

if s is the displacement, how do you find velocity, and how do you find acceleration, in theory?

 

Also, do you remember what to do when you have a function inside a function to derivate? Chain rule?

 

Try to see if oyu can start the derivation, we'll help once we see how far you managed to go.

[Teggle]

Derive the following function to find the velocity and acceleration: s=0.3cos(pi/4*t)

 

 

Chain Rule:

 

s=0.3cos(pi/4*t)

 

Let u= pi/4*t so s= 0.3cosu

 

du/dt= 1*pi/4*t^1-1 ds/du= 1*0.3cosu^1-1

 

du/dt= pi/4 ds/du= 0.3cos

 

 

 

ds/dt= ds/du*du/dt

 

ds/dt= 0.3cos*pi/4

 

ds/dt= 0.3cos pi/4

 

 

However, with this answer "u" disappears. So I believe this is wrong, but I'm not sure. Any more ideas?

 

if s is the displacement, how do you find velocity, and how do you find acceleration, in theory?

 

Also, do you remember what to do when you have a function inside a function to derivate? Chain rule?

 

Try to see if oyu can start the derivation, we'll help once we see how far you managed to go.

 

 

To find velocity, you need to derive s=0.3cos(pi/4*t). To find the accelaration, you need to derive the velocity function. So, in the end, s=0.3cos(pi/4*t) is derived twice to get the answer.

[imatfaal]

Teggle

 

Your differentiation of s = .3 Cos(u) to find ds/du is incorrect. Your layout isn't best in world either - I was always told to use one line for each equation unless there was a good reason to put two on one line - and try and group your ideas more logically, it really helps in the maths to have an easily followed piece of work.

 

Try looking here for the correct way to differentiate Cos(x)

Edited May 1, 2011 by imatfaal

[lackos]

As stated above using the chain rule will allow you to get this problem out.

 

i.e f '(x)=h '(x)*g '(h(x)).

 

Now if you are able to determine your h(x) and g(x) you will find the derivitive.

 

In my opinion this is the best way to approach the chain rule as it is quite clean and methodical.

 

e.g. differentiate f(x)=0.5*(20*x+5)^4 with respect to x

 

here you would assign: g(x)=0.5*x^4

therefore g '(x)= 2*x^3

h(x)=20*x+5

h '(x)=20

therfore f '(x)=20*2*(20*x+5)^3

=40*(20*x+5)^3.

 

apply this too your problem and it should come out.

btw this is my first post and i hope i havent given away too much or too little information.

Edited May 1, 2011 by lackos

[Teggle]

Teggle

 

Your differentiation of s = .3 Cos(u) to find ds/du is incorrect. Your layout isn't best in world either - I was always told to use one line for each equation unless there was a good reason to put two on one line - and try and group your ideas more logically, it really helps in the maths to have an easily followed piece of work.

 

Try looking here for the correct way to differentiate Cos(x)

 

 

I appologise. I didn't realise I had made a mistake. You are correct, I was supposed to make cos(u) into -sin(u). Making the answer: 0.3-sin pi/4 Thank you for pointing this out. As for how I wrote out my last post, when I was typing it up there was more space between my workings out for ds/du and du/dt. When I posted it, however, it took away the large amount of space I had, making it somewhat confusing to read. I appologise for this as well.

 

My question is, if I change cos(u) to -sin(u), does this then make my findings correct? I am still unsure as to whether I am answering this correctly, and I have to derive 0.3-sin pi/4 to fully answer the question (as I have to find the acceleration), and I am unsure of how to do this with my answer. Maybe you could do an example which is similar to what I have to deal with? Thank you.

[Teggle]

Thank you all for helping. I asked my teacher and found out what I was doing wrong.

 

Chain Rule:

 

s=0.3cos(pi/4*t)

 

Let u= pi/4*t

 

du/dt= 1*pi/4*t^1-1

 

du/dt= pi/4

 

 

so s= 0.3cos u

 

ds/du= 0.3-sin u

 

 

 

ds/dt= ds/du*du/dt

 

ds/dt= 0.3-sin u*pi/4

 

ds/dt= 0.3-sin (pi/4*t)*pi/4

 

ds/dt= pi/4*0.3-sin (pi/4*t)

 

 

And then to find the acceleration:

 

v= pi/4*0.3-sin (pi/4*t)

 

Let u= pi/4*t

 

du/dt= pi/4

 

 

so v= pi/4*0.3-sin u

 

dv/du= pi/4*0.3-cos u

 

 

dv/dt= dv/du*du/dt

 

dv/dt= pi/4*0.3-cos u*pi/4

 

dv/dt= pi^2/16*0.3-cos (pi/4*t)

 

 

 

Thanks

[imatfaal]

Teggle - that's cool. To expand on what I meant on layout, (BTW I quite agree it can be a pain to type a nicely aligned piece of work for the browser/printer to reformat it) I will just give you an idea on your last answer

 

dv/dt= pi^2/16*0.3-cos (pi/4*t)

 

1. the minus sign needs to move - whilst in context you can see what's happening, anyone coming to that equation will read it as having two parts

pi^2/16*0.3 LESS cos (pi/4*t)

 

2. in pi^2/16*0.3 - is the .3 multiplying the 16 or the whole thing. similarly is the t multiplying the 4 or the pi/4

 

3. personally I would try and either use decimals or fractions but not both

 

You could use latex on line which is very neat

 

[math]-\frac{3}{160} \pi^2 cos\frac{\pi t}{4}[/math]

 

Or at least rearrange and use brackets to make it clear. The multiplication thing in point 2 might well go away when written by hand - but the minus sign is just bad!

[Teggle]

Teggle - that's cool. To expand on what I meant on layout, (BTW I quite agree it can be a pain to type a nicely aligned piece of work for the browser/printer to reformat it) I will just give you an idea on your last answer

 

dv/dt= pi^2/16*0.3-cos (pi/4*t)

 

1. the minus sign needs to move - whilst in context you can see what's happening, anyone coming to that equation will read it as having two parts

pi^2/16*0.3 LESS cos (pi/4*t)

 

2. in pi^2/16*0.3 - is the .3 multiplying the 16 or the whole thing. similarly is the t multiplying the 4 or the pi/4

 

3. personally I would try and either use decimals or fractions but not both

 

You could use latex on line which is very neat

 

[math]-\frac{3}{160} \pi^2 cos\frac{\pi t}{4}[/math]

 

Or at least rearrange and use brackets to make it clear. The multiplication thing in point 2 might well go away when written by hand - but the minus sign is just bad!

 

 

Thanks for the tips! I'll be sure to use these when writing my assignment!