# Thread: Area of a Triangle Using the Derivative

1. ## Area of a Triangle Using the Derivative

I know this is asking quite a bit, but if you have the free time, I would appreciate your help very much...

On the MIT Opencourseware site, in the single variable calculus class, the professor works a problem that goes roughly like this:

Find the area of a triangle whose sides are formed by the tangent line of a F(x)= 1/x, and the x & y axises. I have provided a link for the video below. My question is this: About the 40:55 minute mark, he comes up with the conclusion that 1/2 *2x0 (x nought)*2y0(y nought) = 2 and I dont understand how he gets 2. If someone has the free time would they mind watching and trying to explain it to me? The problem starts around the 28:15 mark. Again, I know answering this question would prob use up alot of your time, but if you are willing, I would appreciate it so much. Thanks.

MIT OpenCourseWare | Mathematics | 18.01 Single Variable Calculus, Fall 2006 | Video Lectures | Video Lectures - Lecture 1

2. Originally Posted by dbakeg00
I know this is asking quite a bit, but if you have the free time, I would appreciate your help very much...

On the MIT Opencourseware site, in the single variable calculus class, the professor works a problem that goes roughly like this:

Find the area of a triangle whose sides are formed by the tangent line of a F(x)= 1/x, and the x & y axises. I have provided a link for the video below. My question is this: About the 40:55 minute mark, he comes up with the conclusion that 1/2 *2x0 (x nought)*2y0(y nought) = 2 and I dont understand how he gets 2. If someone has the free time would they mind watching and trying to explain it to me? The problem starts around the 28:15 mark. Again, I know answering this question would prob use up alot of your time, but if you are willing, I would appreciate it so much. Thanks.

MIT OpenCourseWare | Mathematics | 18.01 Single Variable Calculus, Fall 2006 | Video Lectures | Video Lectures - Lecture 1

He gets at 41:14 that the area is $2x_0y_0=2$ , and the reason he writes "equal to 2" is that $x_0y_0=1\iff y_0=\frac{1}{x_0}$, which of course is true and

you apparently forgot: we're working on the hyperbola $y=\frac{1}{x}$ and the point $(x_0,y_0)$ is taken from it!

Tonio

3. Originally Posted by tonio
He gets at 41:14 that the area is $2x_0y_0=2$ , and the reason he writes "equal to 2" is that $x_0y_0=1\iff y_0=\frac{1}{x_0}$, which of course is true and

you apparently forgot: we're working on the hyperbola $y=\frac{1}{x}$ and the point $(x_0,y_0)$ is taken from it!

Tonio
Just because of the time factor you mentioned, if it can help save people time: I completely agree with Tonio's response.

4. Originally Posted by undefined
Just because of the time factor you mentioned, if it can help save people time: I completely agree with Tonio's response.

Thanx, but of course I didn't watch the whole video! In fact that looked like a High School calculus class (boooooring), and I just clicked on the relevant times: at minute 28:00 and then at 41:00 .

Tonio

5. Originally Posted by tonio
Thanx, but of course I didn't watch the whole video! In fact that looked like a High School calculus class (boooooring), and I just clicked on the relevant times: at minute 28:00 and then at 41:00 .

Tonio
Me too I probably watched a little over 5 minutes because I wanted to see his teaching style. (Been a while since I've been in school, and I didn't know Opencourseware had online video lectures.)

6. I really didn't put that together. Thanks so much for your help, I really appreciate it!