Talk:Introduction to Calculus/Introduction

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I believe that there is a factual error in the introduction when it states that a falling rock will fall 9.8t^2 meters in t seconds or 16t^2 feet in t seconds. the equation for distance is 1/2 gt^2. in mps^2 this works out to be 4.9t^2 meters in t seconds. does that make sense.

Question I[edit source]

Why can't I continue? There is another heading underneath the lesson on calculating the average speed of a falling rock; however, there is nothing underneath the heading, and there is no obvious link to click to continue. I would like to move on to the next section... Thanks!

It hasn't been written yet. However, there is a wikibook b:Calculus which is probably more useful at this point.--Rayc 07:17, 13 December 2006 (UTC)

Question II[edit source]

What is the exponent above the variable "t" in the equation to determine velocity of the falling rock? Also, why is the exponent 3 for the section asking the student to calculate the velocity of the rocket but 2 for the velocity of a falling rock? Cpuga001 00:19, 3 March 2009 (UTC)

Question III[edit source]

I the example of the falling rock: "a rock will fall at 16t^2 or 9.81t^2 metres" this is very confusing. Acceleration due to gravity is expressed as 9.81 m/s^2, where did the 16 come from and what is it's unit? "the total distance traveled during this interval will be 16t_1^2-16t^2" t^2 is a unit of time not distance.

Its unit is feet. t is meant to be a unit of time, not distance, because the acceleration due to gravity makes distance a function of time.

Question IV[edit source]

I didn't understand why it is 16t² feet or 9.81t²meters. The equation that measure H is H = gt²/2 if the initial velocity s zero( which it is) so it should be 9.81/2 not 9.81x2.

Comment[edit source]

I have found this page long winded. It is full of examples, probably from college textbooks, without getting to the core idea. Calculus is about finding the instantaneous rate verses the average rate. Simple in a nutshell! It is represented by the gradient at a defined point on a curve. Differentiation is about calculating that gradient.