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TonyHem · « **Reply #105 on:** July 05, 2009, 07:15:40 pm »

Dt/Dtheta = 100/theta
T = 100loge(theta)+c
Temperature = 300, when t = 0
0 = 100loge(300)+c
c = -100loge(300)

T = 100loge(theta)-100loge(300)
t=100(loge(theta/300))
10 minutes
10/100 = loge(theta/300)
(e^0.1)x300 = theta

Damo17 · « **Reply #106 on:** July 05, 2009, 07:15:53 pm »

Quote from: Over9000 on July 05, 2009, 07:04:51 pm

If the thermostat in an electric heater failed, the rate of increase of temperature, $\frac{d\theta}{dt}$ , would be $0.01\theta$ degrees per minute where \theta is in degrees Kelvin and t is in minutes. If a heater was switched on at a room temperature of 300K and the thermostat did not function, what would the temperature of the heater be after 10 minutes.
I did this:
$\frac{d\theta}{dt} = 0.01\theta$
$\frac{dt}{d\theta} = \frac{100}{\theta}$
$t= 100log_e(\theta)$
$10 = 100log_e(\theta)$
$0.1 = log_e(\theta)$
$\theta = e^{0.1}$
$\theta = 1.01$
$\therefore 301.1K$
But the answer is 331.55, what did I do wrong, I tried many different ways and got many different answers but I dont see whats wrong with this way.

You forgot the +c after integrating.

$t= 100log_e (\theta) +c$

$t=0$ , $\theta = 300$

$\therefore t= 100log_e (\frac{ \theta }{300})$

$\theta = 300 e^{\frac{t}{100}}$

$t=10$

$\theta = 300e^{0.1} = 331.55$

EDIT: damn, beaten to it.

TrueTears · « **Reply #107 on:** July 05, 2009, 07:25:59 pm »

Always remember the +c >9k that's what I've been telling ya.

Over9000 · « **Reply #108 on:** July 05, 2009, 07:53:26 pm »

Thanks Tonyhem, damo17 and TT. I keep making these small mistakes that will cost me

Over9000 · « **Reply #109 on:** July 28, 2009, 11:32:20 pm »

Find $\frac{dy}{dx}$ by implicit differentiation for this expression: $\frac{x}{e^{2y}} - 5y + 6x^2 + 3 = 0$

The following is my method:

$x - 5e^{2y}y + 6x^2e^{2y} + 3e^{2y} = 0$

$1 - 5 [ 2 \frac{dy}{dx}e^{2y} y + e^{2y} \frac{dy}{dx}] + 6[ 2xe^{2y} + 2x^2 \frac{dy}{dx} e^{2y}] + 6 \frac{dy}{dx} e^{2y} = 0$

$1 - 10 \frac{dy}{dx} e^{2y} y - 5e^{2y} \frac{dy}{dx} + 12x e^{2y} + 12x^2 \frac{dy}{dx} e^{2y} + 6 \frac{dy}{dx} e^{2y} = 0$

$\frac{dy}{dx}(-10e^{2y} y - 5e^{2y} + 12x^2e^{2y} + 6e^{2y}) = -1 - 12xe^{2y}$

$\frac{dy}{dx} = \frac{-1 - 12xe^{2y}}{(-10e^{2y} y - 5e^{2y} + 12x^2e^{2y} + 6e^{2y})}$

$\frac{dy}{dx} = \frac{1+12xe^{2y}}{e^{2y}(10y + 5 - 12x^2 - 6)}$

$\frac{dy}{dx} = \frac{1+12xe^{2y}}{e^{2y}(10y - 12x^2 - 1)}$

But answer has something completely different...

What do I do?

TrueTears · « **Reply #110 on:** July 28, 2009, 11:34:02 pm »

This is similar to /0's question: http://vcenotes.com/forum/index.php/topic,15322.msg164639.html#msg164639

They are the same thing basically just two different methods, you can test this by subbing in the coordinates they should be the same.

/0 · « **Reply #111 on:** July 28, 2009, 11:37:49 pm »

In that particular question I think it's better to not multiply out $e^{2y}$ , cuz more 'y's makes you have to use the chain rule a lot more.

Over9000 · « **Reply #112 on:** October 23, 2009, 08:03:36 pm »

Find the area of the region bounded by the curve with equation y = $\frac{2x-1}{x-2}$ , the horizontal asymptote and the lines x=3 and x=5.
Im stuck??
I just did $\int{\frac{2x-1}{x-2}}dx$ from 3 to 5, but the answer says its $\int{\frac{2x-1}{x-2}}dx$ -4 with same limits.

/0 · « **Reply #113 on:** October 23, 2009, 08:12:51 pm »

$\int_3^5 \frac{2x-1}{x-2}dx$ is the area bounded by the curve, the x-axis, and lines x = 3 and x = 5... not the horizontal asymptote.

Over9000 · « **Reply #114 on:** October 23, 2009, 08:17:43 pm »

Quote from: /0 on October 23, 2009, 08:12:51 pm

$\int_3^5 \frac{2x-1}{x-2}dx$ is the area bounded by the curve, the x-axis, and lines x = 3 and x = 5... not the horizontal asymptote.

The line is y=2, why is it -4 and not +4 or +2?

Wait dw, I think I know why, thanks

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