# Suppose that g(t)...

• November 5th 2009, 12:52 PM
mathcalculushelp
Suppose that g(t)...
Suppose that g(t)=24te^kt for t> 0,where k is some constant.Find an appropriate value of k .
• November 5th 2009, 01:09 PM
tonio
Quote:

Originally Posted by mathcalculushelp
Suppose that g(t)=24te^kt for t> 0,where k is some constant.Find an appropriate value of k .

I think an appropiate value of k could be k = 2, though k = 4 is cutter and k = 345 is, naturally, very handsome.

Don't you people check your own posts??

Tonio
• November 5th 2009, 01:10 PM
e^(i*pi)
Quote:

Originally Posted by mathcalculushelp
Suppose that g(t)=24te^kt for t> 0,where k is some constant.Find an appropriate value of k .

$k = ln(2)$

The question is extremely vague
• November 5th 2009, 01:14 PM
mathcalculushelp
We need to find the value of k for which g has a horizontal tangent line where t=3.
• November 5th 2009, 01:27 PM
e^(i*pi)
Quote:

Originally Posted by tonio
I think an appropiate value of k could be k = 2, though k = 4 is cutter and k = 345 is, naturally, very handsome.

Don't you people check your own posts??

Tonio

Quote:

Originally Posted by mathcalculushelp
We need to find the value of k for which g has a horizontal tangent line where t=3.

That makes more sense :)

We need to find g'(3) in other words and for a horizontal line g'(x) = 0

Use the product rule and the chain rule on g(t)

Spoiler:
$g(t)=24te^{kt}$

$u = 24t \: \rightarrow \: u' = 24$

$v = e^{kt} \: \rightarrow \: v' = ke^{kt}$

$g'(t) = 24e^{kt} + 24kte^{kt} = 24e^{kt}(1+kt)$

$g'(t) = 24e^{kt}(1+kt)$

$g'(3) = 24e^{3k}(1+3k) = 0$

In $\mathbb{R}$ $24e^{3k} > 0$ so it provides no solutions so $1+3k = 0$ which gives $k = -\frac{1}{3}$