is my first derivitive... now i need to prove that my highest point is y=46.8998013
how do I prove it using the 2nd derivitive?
Stacey Warren - Expert brainly.com
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Take your second derivative and set the equation to 0. Those/that is your critical point.
After you get the exact x location of when the second derivative = 0, plug it back into the position equation.
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i have the exact value of x at the highest point it 0.55blah blah.. but how is that proof that its the highest point.. cus ive worked that out not using the second deriv?
The second derivative determines whether the system is concave up or concave down, ie minimum or maximum.
The second derivative determines whether the position graph is concave up or concave down, ie minimum or maximum.
If the second derivative is negative at any given x, that means the position graph is concave down, a maximum. If the second derivative is positive, the position graph is concave up, a minimum.
oooooh so because my second derivative is negative (-392..)... it proves my value of x to be the MAX cus its d2y/dx2 is negative... where if it was a positive value it would have been the min...
Exactly, but take it further.
You found the critical points because those are the MAX and MIN, those show up on the velocity graph. If the velocity fluxes from positive and negative, you can be sure that is the max value on the position graph.
It's hard to explain without graphs to go with it, I'm sorry.
well i have the graph... which coincidently is a representation of a ball travelling through space being acted upon gravity.. which has given me the equation 197.92x^2 + 220.83x 15.108.... so ive defferentiated it as dy/dx = 0 giving me 0 = -395.84x + 220.83 whic gives me the highest points being (0.55,46)... and to me the second derivitive will be d2y/dx2 will be just = -395.84... which as its negative shows it is infact a "sad face" style parabola... which it is... and also as it is a negative value for d2y/dx2 it proves?!?!?! my values of x,y to be the Max point of the curve?!...