Differentiatiation Q (1 Viewer)

[shaon0]

Differentiation Q

Can someone help me differentiate:
P= I/[RI+(1-RI)e^-kt]

 

[vds700]

Re: Differentiation Q

Can someone help me differentiate:
P= I/[RI+(1-RI)e^-kt]

P = I . [RI+(1-RI)e^-kt]^-1

using the chain rule,

dP/dt =I . -[RI+(1-RI)e^-kt]^-2 .(-k)(1-RI)e^-kt
=kI(1-RI)e^-kt/[RI+(1-RI)e^-kt]^2

 

[shaon0]

Re: Differentiation Q

P = [RI+(1-RI)e^-kt]^-1

using the chain rule,

dP/dt = -[RI+(1-RI)e^-kt]^-2 .(-k)(1-RI)e^-kt

its an I not a 1 above the [RI+(1-RI)e^-kt]

 

[vds700]

Re: Differentiation Q

its an I not a 1 above the [RI+(1-RI)e^-kt]

sorryt, i fixed it up

 

[shaon0]

Re: Differentiation Q

sorryt, i fixed it up

Thanks

 

[Fortian09]

Re: Differentiation Q

Hmm I'm having trouble with a differentiation problem

the question is

g(t) = (4t²+7)²(2t³+1)⁴

Find the derivative.

I can get the first bit using chain rule but i cant get the rest..

 

[tommykins]

Re: Differentiation Q

fuck sake. the computer is annoying.

 

[cinDii]

Re: Differentiation Q

are you jus differentiating that?
just go 'first (section) differentiate second (section) + second (section) differentiate first (section) i think the answer is..
g'(t) = 24t^2(2t^3+1)^3(4t^2+7)^2 + 16t(2t^3+1)^4(4t^2+7)
hope thats right and tht helps ><

 

[Fortian09]

Re: Differentiation Q

I have no idea why but both answers are wrong for some reason

 

[tommykins]

Re: Differentiation Q

Nope, my answer is correct. I even graphed it.

 

[Fortian09]

Re: Differentiation Q

Ahh well
man i havent been on this for ages

ummm
I have this question

x² + xy² = 12; (1,3) and ur meant to find the eqns of the tangent and normal to the given curve at hte indicated point

but this question is wierd
I tested the curve by subbing in the point of x=1 but then the point ends up lying on the outside of the curve not the point where the tangent occurs
any answers?

 

[independantz]

Re: Differentiation Q

Ahh well
man i havent been on this for ages

ummm
I have this question

x² + xy² = 12; (1,3) and ur meant to find the eqns of the tangent and normal to the given curve at hte indicated point

but this question is wierd
I tested the curve by subbing in the point of x=1 but then the point ends up lying on the outside of the curve not the point where the tangent occurs
any answers?

Yeah the point does not lie on the curve ??? Probably a dodgy question.

 

[Trebla]

It should be (3,1)

 

[Fortian09]

Re: Differentiation Q

well it's apparently still solvable...
like the point lies on the line of the tangent but isnt actually touching the curve...
so hmmm
iono
it should be solvable i jsut dunno how

 

[Fortian09]

It should be (3,1)

U sure because i tried switching it around and i didnt get the answer either so now wat?

 

[Trebla]

U sure because i tried switching it around and i didnt get the answer either so now wat?

Um...
LHS = x² + xy² sub (3,1)
= 9 + 3.1
= 12
= RHS
so it lies on the curve
So rearranging gives:
12 - x² = xy²
y² = 12/x - x
y = √(12/x - x) (take positive root as y is positive at point (3,1))
Differentiate and you should be able to find equations of tangent and normal
Alternatively you can use implicit differentiation which is an Extension 2 method.

 

[duy.le]

hey tommikins how did u go in 3/4 u? i didnt see any posts by u

 

[tommykins]

i reckon i did pretty well, should be band 6 in both of them.

 

[shaon0]

i reckon i did pretty well, should be band 6 in both of them.

hey tommykins, long time no see. How has the HSC been going?

 

[tommykins]

not too bad, i totally ripped up physics today

english will be the decider for 95uai hahah