Matthew
06-05-2013, 08:53 PM
I'm sorta part of the way there with this question but I can't seem to get what is the correct answer :(
I'm probably just missing something really obvious so if I'm being a moron just say (A)
(BTW, take # to be 'theta')
d#/dt = -k(#+20)
express # in terms of t, k and an arbitary constant.
Okay so I've assumed that this is a 'differential equations' question.
This what I've done:
(Take | to mean 'the integral of')
| (1/#+20) d# = | (-k) dt
so ln(#+20) = -kt + c
#+20 = e^(-kt + c)
# = e^(-kt + c) - 20
-------------------
Now this is NEARLY the correct answer. The correct answer is # = -20 + ce^(-kt)
... so where have I gone wrong? :(
(If anyone can make sense of that, its hard to type this stuff out)
Kardan; Chippiewill; as I know you're good at maths (A)
I'm probably just missing something really obvious so if I'm being a moron just say (A)
(BTW, take # to be 'theta')
d#/dt = -k(#+20)
express # in terms of t, k and an arbitary constant.
Okay so I've assumed that this is a 'differential equations' question.
This what I've done:
(Take | to mean 'the integral of')
| (1/#+20) d# = | (-k) dt
so ln(#+20) = -kt + c
#+20 = e^(-kt + c)
# = e^(-kt + c) - 20
-------------------
Now this is NEARLY the correct answer. The correct answer is # = -20 + ce^(-kt)
... so where have I gone wrong? :(
(If anyone can make sense of that, its hard to type this stuff out)
Kardan; Chippiewill; as I know you're good at maths (A)