## pottersheep 2 years ago I'm confused about rates of change? Why don't I put this into the limit formula? (Grade 12 Adv Functions) Question: As you get father from the earth's surface, gravity has less effect on you. For this reason, you actually weigh less at higher altitudes. A person who weighs 55kg can use this function to find out their weight: [6400(55)]/(h+6400), at a specific height (h) above sea evel. Find the instantaneous rate of change at 12000 ft above sea level. How my teacher told me to do it: [w(12000.001-w(1199.999)]/0.002 = 0.00609 Whatever happened to using LIMITS ?

1. jefftheloveableguy

This Newton's law of universal gravitation i think. Well i guess when taking a limit all you need to do is directly plug in the value into h. This would be the limit as h approached 12000 feet?

2. pottersheep

Isnt this like a "cheat" way to do limits? What happened to properly using the formula?

3. wio

He's not giving you the instantaneous rate of change then, hes giving you the average rate of change over a small interval. If you wanna do a limit you can.

4. jefftheloveableguy

Right i guess so. What he is doing is dw/d(something i dont know) so dw would be a differential and i dont know what the 0.002 is. By the way are you in 12th grade AP physics?

5. jefftheloveableguy

the d of something i dont know would not be the 55 kg because that stays constant

6. pottersheep

Thanks guys