How are graphs calculated?

Hi MHF.

I want to know...If you stand 2 meters above the ground on a platform and kick a ball..it will fly in some sort of equation which ends with +2 which resembles the 2 meter tall platform (graph is in meters).

How do we calculate the coefficient for the squared x and what does it resemble? I never really thought about that one. Also, what is the fundamental reason that a squared number goes in an arc while a cubed number go all kinds of crazy...Why is the power of the x so important in deciding its 'shape'?

Thanks.

Re: How are graphs calculated?

The movement of a ball is a matter of physics, not mathematics. The crucial point is that the downward acceleration, as long as the ball stays "near" the surface of the earth (not tens of miles above) it, is that the acceleration due to gravity remains constant while, ignoring air resistance, there is no acceleration horizontally: the vertical acceleration is -g which is approximately -9.8 m per second per second so the vertical speed, t seconds after the kick, is -9.8t plus the initial speed, so $\displaystyle v_y(t)= -gt+ v_{y0}$, while the horizontal velocity remains the initial horizontal speed: $\displaystyle v_x(t)= v_{x0}$. That means that, in t seconds, the ball goes, horizontally, distance $\displaystyle x= v_{x0}t$. Vertically, with changing speed, it's a little harder but with **constant** acceleration, we can use the **average** speed. With starting speed, vertically, $\displaystyle v_{y0}$, and speed after t seconds, $\displaystyle v_{y0}- gt$, the average speed will be $\displaystyle \frac{v_{y0}+ v_{y0}+ gt}{2}= v_{y0}- (g/2)t$ and so the ball have gone, vertically, the distance $\displaystyle \left(v_{y0}- (g/2)t\right)t= -(g/2)t^2+ v_{y0}t$.

That is, $\displaystyle x= v_{x0}t$ while $\displaystyle y= -(g/2)t^2+ v_{y0}t$. To get that in terms of x and y only, from $\displaystyle x= v_{x0}t$ we get $\displaystyle t= \frac{x}{v_x0}$ and then $\displaystyle y= -(g/2)(x/v_{x0})^2+ v_{y0}(x/v_0)= -\frac{g}{2v_{x0}^2}x^2+ \frac{v_{y0}}{v_{x0}}t$.

That's what the "coefficient for the squared x" is- it depends upon the acceleration due to gravity and the initial horizontal speed. Yes the graph of "a squared number" goes in an arc- more correctly, a parabola. But I would not say that a cubed number goes "all kinds of crazy"- it just has **two** arcs. You probably know that a quadratic equation, $\displaystyle x^2+ bx+ c= 0$ may have **two** solutions so its graph has to cross y= 0 twice. Similarly, an equation involving $\displaystyle x^3$ may have three roots and so its graph may cross y= 0 **three times**. If it goes up at, say x= 0, it must reach some maximum value and then come back down to cross the axis again, the turn and cross a third time- two arcs instead of one.

Re: How are graphs calculated?

Quote:

Originally Posted by

**HallsofIvy** The movement of a ball is a matter of physics, not mathematics. The crucial point is that the downward acceleration, as long as the ball stays "near" the surface of the earth (not tens of miles above) it, is that the acceleration due to gravity remains constant while, ignoring air resistance, there is no acceleration horizontally: the vertical acceleration is -g which is approximately -9.8 m per second per second so the vertical speed, t seconds after the kick, is -9.8t plus the initial speed, so $\displaystyle v_y(t)= -gt+ v_{y0}$, while the horizontal velocity remains the initial horizontal speed: $\displaystyle v_x(t)= v_{x0}$. That means that, in t seconds, the ball goes, horizontally, distance $\displaystyle x= v_{x0}t$. Vertically, with changing speed, it's a little harder but with **constant** acceleration, we can use the **average** speed. With starting speed, vertically, $\displaystyle v_{y0}$, and speed after t seconds, $\displaystyle v_{y0}- gt$, the average speed will be $\displaystyle \frac{v_{y0}+ v_{y0}+ gt}{2}= v_{y0}- (g/2)t$ and so the ball have gone, vertically, the distance $\displaystyle \left(v_{y0}- (g/2)t\right)t= -(g/2)t^2+ v_{y0}t$.

That is, $\displaystyle x= v_{x0}t$ while $\displaystyle y= -(g/2)t^2+ v_{y0}t$. To get that in terms of x and y only, from $\displaystyle x= v_{x0}t$ we get $\displaystyle t= \frac{x}{v_x0}$ and then $\displaystyle y= -(g/2)(x/v_{x0})^2+ v_{y0}(x/v_0)= -\frac{g}{2v_{x0}^2}x^2+ \frac{v_{y0}}{v_{x0}}t$.

That's what the "coefficient for the squared x" is- it depends upon the acceleration due to gravity and the initial horizontal speed. Yes the graph of "a squared number" goes in an arc- more correctly, a parabola. But I would not say that a cubed number goes "all kinds of crazy"- it just has **two** arcs. You probably know that a quadratic equation, $\displaystyle x^2+ bx+ c= 0$ may have **two** solutions so its graph has to cross y= 0 twice. Similarly, an equation involving $\displaystyle x^3$ may have three roots and so its graph may cross y= 0 **three times**. If it goes up at, say x= 0, it must reach some maximum value and then come back down to cross the axis again, the turn and cross a third time- two arcs instead of one.

This is very interesting. I can't wait to learn physics!

Thank you for your answer! If I may ask...If I have the function y=x^3...Does the curve go to infinity...And then back? Thanks.