C4 differentiation help
I think I've differentiated these two equations wrong, so I keep getting my final answer wrong! Help would be much appreciated!
x = sec(t)
y = tan(t)
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x = sec(t)
y = tan(t)
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Original post by kendellex
I think I've differentiated these two equations wrong, so I keep getting my final answer wrong! Help would be much appreciated!
x = sec(t)
y = tan(t)
x = sec(t)
y = tan(t)
Sure, if you show your working I will see where you may be going wrong
Original post by kendellex
I think I've differentiated these two equations wrong, so I keep getting my final answer wrong! Help would be much appreciated!
x = sec(t)
y = tan(t)
x = sec(t)
y = tan(t)
for x, try rewriting sec(t) in terms of functions you're used to... then differentiate using the quotient or product rule.
do the same for y, and see what you get.
if you can't do it, there are standard derivatives in the formula booklet, but you should be able to prove them.
Original post by TenOfThem
Sure, if you show your working I will see where you may be going wrong
I thought this.
dx/dt = sec(t)tan(t)
dy/dt = sec^2(t)
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Original post by Mike_123
for x, try rewriting sec(t) in terms of functions you're used to... then differentiate using the quotient or product rule.
do the same for y, and see what you get.
if you can't do it, there are standard derivatives in the formula booklet, but you should be able to prove them.
do the same for y, and see what you get.
if you can't do it, there are standard derivatives in the formula booklet, but you should be able to prove them.
I don't actually have a formulae sheet or a copy of C3
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Original post by kendellex
So far so good
Original post by kendellex
And you get (chain rule)
Original post by TenOfThem
So far so good
Okay, thanks!
I'll try to find out dy/dx again.
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Original post by kendellex
Perfectly correct.
Original post by kendellex
Both are essential - formula book should never leave your hand. Most boards have the spec and formulas to download.
Thank you (:
Original post by Old_Simon
Both are essential - formula book should never leave your hand. Most boards have the spec and formulas to download.
I've never had them :P always have to flick through the book to find formula
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Original post by kendellex
Thank you (:
I've never had them :P always have to flick through the book to find formula
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I've never had them :P always have to flick through the book to find formula
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well the value in using the formula book through your course is that you get used to using it, you learn what is in there, (and just as important what is not in there). They are available online.
Original post by Old_Simon
well the value in using the formula book through your course is that you get used to using it, you learn what is in there, (and just as important what is not in there). They are available online.
I think ill print one off tomorrow, although I am currently creating one in the front of my book for handiness
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Also here is another question which I think I got wrong :P
x = 2t - sin2t
dx/dt = 2 - t(cos2t)
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x = 2t - sin2t
dx/dt = 2 - t(cos2t)
Posted from TSR Mobile
Original post by kendellex
I think ill print one off tomorrow, although I am currently creating one in the front of my book for handiness
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The thing is people treat them as exam tools. But really they are study tools. The first thing to think of is "Is the formula in the book" ? Each module only has a very few formula you need learn. It is no good teachers saying you need to learn it all etc. That is how the modules work and you are entitled to use all the tools available. In fact you will end up remembering them all anyway but that is not the point. There are so many questions in this very forum from peeps not using the formula books. You can tell by the questions.
Original post by kendellex
Also here is another question which I think I got wrong :P
x = 2t - sin2t
dx/dt = 2 - t(cos2t)
x = 2t - sin2t
dx/dt = 2 - t(cos2t)
where did that come from
Original post by TenOfThem
where did that come from
Absolutely no idea.
I thought if you differentiated (2t) you get 2.
Then if you differentiated (-sin2t) you get -t(cos2t). Or is it -2(cos2t). Or neither? :P
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dy/dx = (dy/dt)/(dx/dt).
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Original post by kendellex
Absolutely no idea.
I thought if you differentiated (2t) you get 2.
Then if you differentiated (-sin2t) you get -t(cos2t). Or is it -2(cos2t). Or neither? :P
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I thought if you differentiated (2t) you get 2.
Then if you differentiated (-sin2t) you get -t(cos2t). Or is it -2(cos2t). Or neither? :P
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This
Since you are using the chain rule
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