C3 differentiation question

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Can someone help me with part a of this question. They said A is a constant in the question so how I went about working it out was as followed:

I read the graph at where t = 0 and saw that T = 12, from this I formulated an equation to get
12 = 5 + Ae^-0
7 = A

This isnt how the markscheme went about it, but I ended up with a different answer to them, but I dont understand why it shouldnt work since A is a constant.

I then did

18 = 5 + 7e^-10k
13/7 = e^-10k
e^10k = 7/13
10k = Ln (7/13)
k = [Ln (7/13)] / 10
Which gives me an answer of -0.0619039....

The actual answer is

Spoiler

Posted from TSR Mobile

(edited 8 years ago)

Reply 1

lizard54142

6

Original post by Hudl

Can someone help me with part a of this question

Posted from TSR Mobile

Where are you stuck? Part (a) is just simultaneous equations.

Reply 2

Theafricanlegend

15

hey you alright mate?

right sub in the values you have for T and t. they must be the corresponding values otherwise it wont work

do that for the two sets of values you have. You'll get simultaenous equations. which hopefully you can solve.

give me a shout if you need further guidance

Reply 3

Hudl

OP

16

Original post by lizard54142

Where are you stuck? Part (a) is just simultaneous equations.

Original post by Theafricanlegend

hey you alright mate?

right sub in the values you have for T and t. they must be the corresponding values otherwise it wont work

do that for the two sets of values you have. You'll get simultaenous equations. which hopefully you can solve.

give me a shout if you need further guidance

Thanks guys, sorry I took the pic on my phone but typed up on computer. If you don't mind scrolling up & I know its simultaneous equations

Reply 4

RhymeAsylumForever

15

Original post by Hudl

Can someone help me with part a of this question. They said A is a constant in the question so how I went about working it out was as followed:

I read the graph at where t = 0 and saw that T = 12, from this I formulated an equation to get
12 = 5 + Ae^-0
7 = A

This isnt how the markscheme went about it, but I ended up with a different answer to them, but I dont understand why it shouldnt work since A is a constant.

I then did

18 = 5 + 7e^-10k
13/7 = e^-10k
e^10k = 7/13
10k = Ln (7/13)
k = [Ln (7/13)] / 10
Which gives me an answer of -0.0619039....

The actual answer is

Spoiler

Posted from TSR Mobile

I did this question earlier today, I'll upload a picture, hold up

Reply 5

Remedium

6

Using
18 = 5 + Ae^-10k

Write an equation for A in terms of k

A = 13/(e^-10k)
So A = 13e^-10k

Substitute this into the other equation and solve:

(7/13) = e^-60k(e^-10k)

(7/13) = e^-50k

etc. to get the value of k

(edited 8 years ago)

Reply 6

lizard54142

6

Original post by Hudl

Can someone help me with part a of this question. They said A is a constant in the question so how I went about working it out was as followed:

I read the graph at where t = 0 and saw that T = 12, from this I formulated an equation to get
12 = 5 + Ae^-0
7 = A

This isnt how the markscheme went about it, but I ended up with a different answer to them, but I dont understand why it shouldnt work since A is a constant.

I then did

18 = 5 + 7e^-10k
13/7 = e^-10k
e^10k = 7/13
10k = Ln (7/13)
k = [Ln (7/13)] / 10
Which gives me an answer of -0.0619039....

The actual answer is

Spoiler

Posted from TSR Mobile

The equation given is only valid for 10<=t<=60, so you cannot use a graphical method to find A; it must be done simultaneously.

Reply 7

Theafricanlegend

15

yeah you cant do that. Your intial equation is wrong. you can't read off the graph at t=0 then sub in T=12 in the generaly equation given. The general equation only works between 10 and 60. You need to do simultaneous equations. lazy m9

Reply 8

Hudl

OP

16

Original post by lizard54142

The equation given is only valid for 10<=t<=60, so you cannot use a graphical method to find A; it must be done simultaneously.

Thanks a lot, I wish I could rep, but Ive repped u today it wont let me & I didnt read that, I skipped all the waffle, turned out to be important for once