Cal 1 problem..

[Vince]

Is it just a matter of pluging in? Not sure what to do...

thanks.

[Brandon13]

Re: Cal 1 problem..

Looks like OS doesn't have any math majors.

[The GIGGAS]

Re: Cal 1 problem..

Looks like OS doesn't have any math majors.

LMAO... what a stupid major.

Then again, that's how I feel.

Did you plug in both numbers and take the difference between them?

[Vince]

Re: Cal 1 problem..

Meh, doesn't matter, i'll just ask my professor tomorrow.

It's a webwork type homework so It was due a few hours ago.

[JohnnytheSkin]

Re: Cal 1 problem..

Rate of change is the derivative of the function with respect to time. You can THEN plug in the different days to find the average rate of change (I trust you can calculate an average).

The instantaneous rate of change is the slope of the tangent line, or the derivative of the function, at a specific point.

[TJdaSportsGuy]

Re: Cal 1 problem..

Rate of change is the derivative of the function with respect to time. You can THEN plug in the different days to find the average rate of change (I trust you can calculate an average).

The instantaneous rate of change is the slope of the tangent line, or the derivative of the function, at a specific point.

Huh? My head just almost exploded when I read all that. Thank GOD I'm done with school...

[The GIGGAS]

Re: Cal 1 problem..

Mine did too... but only because we aren't at derivatives yet.

My damn professor has been going over limits for what... 4 weeks now?

[TJdaSportsGuy]

Re: Cal 1 problem..

Mine did too... but only because we aren't at derivatives yet.

My damn professor has been going over limits for what... 4 weeks now?

Limits...okay. I got that. As in I'm limited in the amount of calculus I can look at before my head explodes. I hate math.

[Vince]

Re: Cal 1 problem..

Mine did too... but only because we aren't at derivatives yet.

My damn professor has been going over limits for what... 4 weeks now?

I hate limits.

[Vince]

Re: Cal 1 problem..

Rate of change is the derivative of the function with respect to time. You can THEN plug in the different days to find the average rate of change (I trust you can calculate an average).

The instantaneous rate of change is the slope of the tangent line, or the derivative of the function, at a specific point.

Thanks buddy

[dkgojackets]

Re: Cal 1 problem..

Seems like you should use quotient rule to take the derivative, then plug the first number in and subtract it from what you get when you plug the second number in. Thats your change in rate.

[TJdaSportsGuy]

Re: Cal 1 problem..

Seems like you should use quotient rule to take the derivative, then plug the first number in and subtract it from what you get when you plug the second number in. Thats your change in rate.

STOP!!

[bkrich83]

Re: Cal 1 problem..

Rate of change is the derivative of the function with respect to time. You can THEN plug in the different days to find the average rate of change (I trust you can calculate an average).

The instantaneous rate of change is the slope of the tangent line, or the derivative of the function, at a specific point.

Yup, you went to School of Mines alright.

[JohnnytheSkin]

Re: Cal 1 problem..

Yup, you went to School of Mines alright.

Don't you mean the "Colorado School of MINDS"?

[BGarrett7]

Re: Cal 1 problem..

LMAO... what a stupid major.

Lame.

Disappointed I missed out on this thread.

Agree with dk though, definitely quotient rule.