To get this much of knowlwdge for free, you re like a guardian angel to me.And not only mine,also to the whole A level students of Nepal. Thanks Mr. Stuart
thank u and it was an excellent explaination i like to watch some more videos regarding maths because it was easy in understanding ..................................
hello parametric equations x=f(t),y=(t) where y is also a differentiable funtion of x then dy/dx=(dy/dt)/(dx/dt) what if i can't eliminate the parameter t (to express the parametric curve in terms of x and y) , can y still be treating as a function of x??? for example x=t*exp(t) , y=t*sin(t) how do you say that y is a function of x [y=h(x)] ? the term dt/dx be used when t is considered as a function of x,but what if I can't express t in terms of x The formula can still be used?
I haven't watched the whole video yet so idk what part is this but if it's simply just sint/3cost it would be 1/3tant because sint/3cost is the same as saying 1sint/3cost. So if it was 3sint/1cost it would be 3tant. If it was 3sint/3cost it would be tant. Edit: Oh the part you're talking about, -sint/3cost would be -1/3tant
it's not spelled with a z as far as I am aware so being pedantic the awnser to your question is no but factorise is definately a word in the oxford dictionary.
I'm massively stuck on one question at the moment: x= (2e^t)-5 y=(8e^-t)+(3e^t)-4 Find the value of t when dy/dx =1 I got dy/de to be: (-8te^-t-1)+(3te^t-1) and de/dx to be: 1/(2te^t-1) Putting these together gives me dy/dx = [(-8te^-t-1)/(2te^t-1)]+[(3te^t-1)/(2te^t-1)] Ultimately, simplifying this and rearranging for dy/dx =-1 I get -1 = (-4e^-2t) + (3/2) By that is apparent ally impossible to solve. Can anyone help?
I would've never got an A* in my exam if it weren't for your amazing videos! Thank you so much
You're welcome and well done on getting an A*.
Hi
@@danleprince7856😂😂😂
The 2 from the half does not go up to the top. It is 1/2 multiplied by (-3) which gives -3/2.
To get this much of knowlwdge for free, you re like a guardian angel to me.And not only mine,also to the whole A level students of Nepal. Thanks Mr. Stuart
You are more than welcome
ماشاء الله بارك الله فيك ❤❤
Thanks man! I like how you can explain this so much better than my actual lecturer in uni
+addy583 No probs. Pleased to help out.
Hello:) I am wonderin, when you do the differentiation of -1/2t^3, doesn't it come out with 6t^-4 which also=6/t^4?
Y doesnt the two go up since it is the coefficient of t
5:11 why is it not -4sint
thanks this was to the point and extremely useful
thank u
and it was an excellent explaination
i like to watch some more videos regarding maths
because it was easy in understanding ..................................
hello
parametric equations x=f(t),y=(t)
where y is also a differentiable funtion of x
then dy/dx=(dy/dt)/(dx/dt)
what if i can't eliminate the parameter t (to express the parametric curve in terms of x and y) , can y still be treating as a function of x???
for example x=t*exp(t) , y=t*sin(t)
how do you say that y is a function of x [y=h(x)] ?
the term dt/dx be used when t is considered as a function of x,but what if I can't express t in terms of x
The formula can still be used?
Thanks sir, I have learnt too much from your videos
❤
@MrLiverwurst As far as I am aware yes. Americans tend to say factor out.
I got 20/20 on my exam thank you you ❤❤
u the best bro
hey man thanks a lot , but how do I know which one divides the other? I really suck at math
does the sint/3cost not cancel out to 3tant? or can you not do that?
I haven't watched the whole video yet so idk what part is this but if it's simply just sint/3cost it would be 1/3tant because sint/3cost is the same as saying 1sint/3cost. So if it was 3sint/1cost it would be 3tant. If it was 3sint/3cost it would be tant.
Edit: Oh the part you're talking about, -sint/3cost would be -1/3tant
thanks this was great
You are the best!! so in this topic we don't apply product rule?
I did when I found dy/dx. dy/dx = dy/dt x dt/dx = [dy/dt] / [dx/dt]
i'm enjoy with this video
You are great ! thanks a ton
thanks... nice job mate
for the differentiation of x=3sin(t) why isn't the product rule used? Am I missing something?
+MilahMee Because we are not multiplying two variables together. 3 is a constant.
it's not spelled with a z as far as I am aware so being pedantic the awnser to your question is no but factorise is definately a word in the oxford dictionary.
you spelt 'answer' and 'definitely' wrong and the z spelling is the American way, so technically it's still correct.
I'm massively stuck on one question at the moment:
x= (2e^t)-5
y=(8e^-t)+(3e^t)-4
Find the value of t when dy/dx =1
I got dy/de to be: (-8te^-t-1)+(3te^t-1)
and de/dx to be: 1/(2te^t-1)
Putting these together gives me
dy/dx = [(-8te^-t-1)/(2te^t-1)]+[(3te^t-1)/(2te^t-1)]
Ultimately, simplifying this and rearranging for dy/dx =-1 I get
-1 = (-4e^-2t) + (3/2)
By that is apparent ally impossible to solve.
Can anyone help?
+lolzomgz1337 once try with logarithimic differrentiation if not then u can use parametric
X=sin2thita y=cos3thita evaluate dy/dx+3/2 Y^1/3
thank you sir👍
Appreciated - thanks
great vid
y=2-3t/1+t, x=3+2t/1+t can you help me with this?
To save time I just do y dot / x dot
It would cancel to (1/3)tant
Would -sint/3cost not cancel to -1/3tant?
Too complicated to do.