Rates-Change with lns and sin/cos

**SportfreundeKeaneKent** · May 27th 2007, 11:52 AM

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

**Jhevon** · May 27th 2007, 12:12 PM

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 $S(12)$

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

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

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

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

Do you know how to find the derivative of ln ?

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

This means the original score is 72

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

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

we want $t$ such that:

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

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

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

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

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

**Jhevon** · May 27th 2007, 12:25 PM

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 $v'(t)$

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

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

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

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

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

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

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

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

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

EDIT 1: added $\pi$ and $2 \pi$ 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?

**behemoth100** · May 27th 2007, 01:06 PM

Originally Posted by Jhevon

The second

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

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

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

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

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

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

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

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

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

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

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).
So isn't your answer incomplete?

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

**Jhevon** · May 27th 2007, 01:08 PM

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).
So isn't your answer incomplete?

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

"almighty Jhevon"? where did that come from?

You are correct sir! my solution is incomplete, I shall add your answers to mine

**SportfreundeKeaneKent** · May 27th 2007, 03:48 PM

Nah, I wouldn't have caught that, I'm new to this stuff. Thanks anyways, with the first one.

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