Thread: Suppose that g(t)...

1. Suppose that g(t)...

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

2. 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

3. Originally Posted by mathcalculushelp
Suppose that g(t)=24te^kt for t> 0,where k is some constant.Find an appropriate value of k .
$\displaystyle k = ln(2)$

The question is extremely vague

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

5. 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
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:
$\displaystyle g(t)=24te^{kt}$

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

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

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

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

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

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