Integration & Differentiation Question!!!?

Question

dN/dt = K(50-t)/(2e^(-0.02t)+3t)
let v = 2+3t(e^(-0.02t)) find, dv/dt.
if K=4,
express N in terms of t.
(How to do this???)

Answer 1

I suspect there's a typo in there somewhere. It looks as if it's supposed to work out so that you get dN/dt is something nice in terms of v and dv/dt, and then you can derive N in terms of v (and thence in terms of t). However, this doesn't work in this case.

v = 2 + 3t e^(-0.02t)
dv/dt = 3 e^(-0.02t) + 3t e^(-0.02t) (-0.02)
= 3 e^(-0.02t) (1 - 0.02 t)
= (3/50) e^(-0.02t) (50 - t)

dN/dt = 4 (50-t) / (2e^(-0.02t) + 3t)
= [4(50/3) e^(0.02t) (dv/dt)] / [e^(-0.02t) (2 + 3te^(0.02t)) ]

Note that the second factor in the denominator is similar to v, but we have e^0.02t instead of e^-0.02t.

I suggest that the formula for dN/dt should be
dN/dt = K(50-t)/(2e^(0.02t)+3t)
Then we could express this as
dN/dt = [4(50/3) e^(0.02t) (dv/dt)] / [e^(0.02t) (2 + 3te^(-0.02t)) ]
= 200/3 (dv/dt) / v
whence we get N = 200/3 (ln |v|) + const
=> N = 200/3 ln |2 + 3t e^(-0.02t)| + const.

If the problem is indeed as stated, we get
dN/dt = [4(50/3) e^(0.02t) (dv/dt)] / [e^(0.02t) (2e^(-0.04t) + 3te^(-0.02t)) ]
= [200/3 dv/dt] / [v + 2e^(-0.04t) - 2]
I don't think there's much you can do with this.

Answer 2

12. make f(x) = y to do implicit, then take organic log of both side: lny = ln(3x^3 + 4)^(x + a million), use organic log regulations: lny = (x+a million)ln(3x^3+4), take by-product(remember chain rule): (a million/y)dy = (ln(3x^3+4)+((9x^2)(x+a million))/(3x^3+4))dx, multiply both side through y (or extremely through what y equalled initially): dy/dx = (ln(3x^3+4)+((9x^2)(x+a million))/(3x^3+4))(3x^3 + 4 )^(x+a million), plug in a million for x: f'(a million) = 14ln7+36 ~ sixty 3.243