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Mathematics 22 Online
OpenStudy (anonymous):

Calculus. s=(t^3)/3 -6t^2 +50t with what value of t does the object have a minimum velocity.

OpenStudy (anonymous):

ive worked the velocity equation out by differentiating the equation, and have got t^2 -12t +50

OpenStudy (anonymous):

so then t^2 -12t +50 =0

OpenStudy (anonymous):

i cant factorise this to find the points by which the gradient = 0

OpenStudy (turingtest):

then use the quadratic formula

OpenStudy (turingtest):

in case you forgot\[ax^2+bx+c=0\implies x={-b\pm\sqrt{b^2-4ac}\over2a}\]

OpenStudy (anonymous):

ah ok, havent used that in ages, cheers mate

OpenStudy (anonymous):

i need to get 6

OpenStudy (anonymous):

and using this formula i just get a mathematics error

OpenStudy (turingtest):

oh I didn't read your Q carefully enough, my bad...

OpenStudy (anonymous):

no worries mate

OpenStudy (turingtest):

you are given a function for position, and you want to minimize velocity velocity is the first derivative of position to find max/mins for the velocity function you have to take the derivative again, and set \(that\) equal to zero

OpenStudy (anonymous):

why do you take the second derivative? the second derivative is the derivative of a quadratic

OpenStudy (anonymous):

thus being linear

OpenStudy (turingtest):

to find the max/mins of any function take the derivative and set it to zero

OpenStudy (anonymous):

could you please explain this to me, the second derivative is meant to give a point of inflection or tell you the nature of a turning point, so says my maths teacher

OpenStudy (anonymous):

but when i set the equation to 0 i get an equation that is unfactorable

OpenStudy (turingtest):

you are given position, derivative of position is velocity to find the max/min of the velocity function though you must set the derivative of *that function* equal to zero yes setting the second derivative of the position function equal to zero will also tell you a point of inflection, but for the *position* function, not the velocity one you already have the velocity function, so take the derivative of that and set it to zero

OpenStudy (anonymous):

ah yes i see now, so taking the derivative of the velocity function will give you the maximum/minimum of it. But since the velocity function is a parabola it will only give you a max or min

OpenStudy (anonymous):

Cheers for that mate ;)

OpenStudy (turingtest):

solving s(t)=0 tells you when position is zer solving s'(t)=0 tells you the max/mins of the position and when the velocity is zero we don't want that, we want to know the max/mins of the velocity function so solve s''(t)=0 you're welcome (from across the pond I assume)!

OpenStudy (anonymous):

haha yes, guess you can tell where then haha.

OpenStudy (anonymous):

judging by the term im assuming your british?

OpenStudy (turingtest):

no, but my stepfather is so I'm quite familiar with British English I'm a US citizen living in Mexico

OpenStudy (anonymous):

Ah, interesting. Well Aussie here mate, haha, cheers for the help. Not familiar with the time zones, but i believe its day time over there, so have a good day mate.

OpenStudy (turingtest):

ah, then you're more than the "pond" away :P see ya around!

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