# Thread: Rates-Change with lns and sin/cos

1. ## Rates-Change with lns and sin/cos

I can't get the right answer for these two, mainly because they include trigonometry and lns.

2. Originally Posted by SportfreundeKeaneKent
I can't get the right answer for these two, mainly because they include trigonometry and lns.

For the first:

(a)
we want $\displaystyle S(12)$

$\displaystyle S(12) = 72 - 15 \ln (12 + 1) = 72 - 15 \ln (13) \approx 33.53$

(b)
we want $\displaystyle S'(4)$

Now $\displaystyle S'(t) = - \frac {15}{t + 1}$

$\displaystyle \Rightarrow S'(4) = - \frac {15}{5} = -3$

Do you know how to find the derivative of ln ?

(c)
I assume the original score is to be taken when $\displaystyle t = 0$

This means the original score is 72

we want to know when does the average score becomes less than $\displaystyle \frac {3}{4} 72 = 54$

So we must solve for $\displaystyle S(t) < 54$

we want $\displaystyle t$ such that:

$\displaystyle S(t) = 72 - 15 \ln(t + 1) < 54$

$\displaystyle \Rightarrow -15 \ln(t + 1) < -18$

$\displaystyle \Rightarrow ln(t + 1) > \frac {6}{5}$

$\displaystyle \Rightarrow e^{ \frac {6}{5}} > t + 1$

$\displaystyle \Rightarrow t > e^{ \frac {6}{5}} - 1$

3. Originally Posted by SportfreundeKeaneKent
I can't get the right answer for these two, mainly because they include trigonometry and lns.

The second

The rate of change of voltage is $\displaystyle v'(t)$

Now, $\displaystyle v'(t) = -2 \sin(t) - 2 \sin(2t)$ ........By the Chain rule

we want $\displaystyle t$ such that $\displaystyle v'(t) = 0$

That is, we want to solve $\displaystyle -2 \sin(t) - 2 \sin(2t) = 0$

$\displaystyle \Rightarrow \sin(t) + \sin(2t) = 0$

$\displaystyle \Rightarrow \sin(t) + 2 \sin(t) \cos(t) = 0$

$\displaystyle \Rightarrow \sin(t) (1 + 2 \cos(t)) = 0$

$\displaystyle \Rightarrow \sin(t) = 0 \mbox { or} 1 + 2 \cos(t) = 0$

$\displaystyle \Rightarrow t = 0 \mbox { or} \cos(t) = - \frac {1}{2}$

$\displaystyle \Rightarrow t = 0 \mbox { , } \pi \mbox { , } 2 \pi \mbox { or } t = \frac {2 \pi}{3} \mbox { , } \frac {4 \pi}{3}$

EDIT 1: added $\displaystyle \pi$ and $\displaystyle 2 \pi$ to the solutions

EDIT 2: Thanks for looking out behemoth100...would you believe that I intentionally left the answer incomplete to see if SportfreundeKeaneKent would catch it?

4. Originally Posted by Jhevon
The second

The rate of change of voltage is $\displaystyle v'(t)$

Now, $\displaystyle v'(t) = -2 \sin(t) - 2 \sin(2t)$ ........By the Chain rule

we want $\displaystyle t$ such that $\displaystyle v'(t) = 0$

That is, we want to solve $\displaystyle -2 \sin(t) - 2 \sin(2t) = 0$

$\displaystyle \Rightarrow \sin(t) + \sin(2t) = 0$

$\displaystyle \Rightarrow \sin(t) + 2 \sin(t) \cos(t) = 0$

$\displaystyle \Rightarrow \sin(t) (1 + 2 \cos(t)) = 0$

$\displaystyle \Rightarrow \sin(t) = 0 \mbox { or} 1 + 2 \cos(t) = 0$

$\displaystyle \Rightarrow t = 0 \mbox { or} \cos(t) = - \frac {1}{2}$

$\displaystyle \Rightarrow t = 0 \mbox { or } t = \frac {2 \pi}{3} \mbox { , } \frac {4 \pi}{3}$
Ok well given that its only 7am in the morning this may be a stupid comment but...
Isnt sin(t) = 0 at 0 pi and 2pi? The interval is 0<t<2pi (except they are equal to or less than signs).

If I have made a mistake on BASIC DIFFERENTIATION AND TRIG THEN I SHOULD BE SHOT!!! I'm scared agaist going up against the almighty Jhevon

5. Originally Posted by behemoth100
Ok well given that its only 7am in the morning this may be a stupid comment but...
Isnt sin(t) = 0 at 0 pi and 2pi? The interval is 0<t<2pi (except they are equal to or less than signs).