Is anyone all right at calculus 1?
Seriously...I need to make sure my math is right and that I'm saying the right thing.... I doubt anyone will help me though, but I would be super grateful if someone could :(
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amelia25 · 26-30, F
Post a picture of it so I can see whether or not I remember how to do it. No guarantees lol
ShaythePanTransMan · 22-25, T
@amelia25 Well, it's an essay. This is the part: "A problem that may arise in a physics engine is as follows: “The formula h(t) = -16t2 + 32t + 80 gives the height h above ground, in feet, of [a projectile] thrown, at t = 0, straight upward from the top of an 80 feet building. What is the highest point reached by the object" (“Projectile Problems with Solutions”)? A well-programmed physics engine would easily figure out this problem. To find the maximum of the height function, the program needs to find the critical numbers of the derivative of the height function. In the language of calculus, the velocity function would be h’(t). To solve the problem, the program would take the derivative of the height function, which is h’(t) = -32t + 32. Then, the program would have to find the critical numbers, or zeros, of that derivative by setting the derivative to 0 and solving for t, which means t = 1. Now, the program plugs that 1 into t in the original equation to find the y value, which is 96. This means the...."
But I'm not sure what it means lol we just learned this recently and I forgot it already
But I'm not sure what it means lol we just learned this recently and I forgot it already
BeefySenpie · M
@ShaythePanTransMan So your max height is 96ft but you might need to do a second derivative or sign test to prove it's a maximum not a minimum
ShaythePanTransMan · 22-25, T
@BeefySenpie Oh...yes, I was looking for the max height. It must be a maximum though; it’s a large number?
BlueVeins · 22-25
@ShaythePanTransMan The math is correct. When you say you're looking for the t value at which v'(t) = 0, what that really means is that you're looking for the moment in time at which the projectile is neither rising nor falling. In practical terms, this is the top of the projectile's arch. As @BeefySenpie pointed out, it's technically necessary to determine the signs of the velocities directly before and after this leveling off [i]in the general case[/i], as a projectile could theoretically do the opposite if an upward force were being applied, but one could reasonably argue that this isn't necessary in the case of this projectile because the only forces being applied are air resistance, gravity, and the force of the throw.
