Naked Science Forum
General Science => General Science => Topic started by: Wasee on 12/01/2009 08:41:43
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Wasee asked the Naked Scientists:
Hello,
I know what derivatives are and how to find the derivative of a function, but what bothers me is how they are used to find out rates of change velocity, accelaration etc
Thanks
What do you think?
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distance (differentiate) > velocity (differentiate) > acceleration
Draw the graph and then its derivative and you'll see it as the gradient function. The x-intercepts of the gradient function are the turning point(s) of the original graph.
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If I was to draw a diagram, like so:
[diagram=391_0]
The co-ordinates of P are (x, (f(x)). If the x-coordinate of Q is (x+h) then the y-coordinate of Q is f(x+h). Gradient of PQ = QR/PR
= (f(x+h) - f(x))/h
As Q 'slides down' the curve: In the limit, as Q→P, and h→0, the gradient of the chord PQ approaches the gradient of the tangent at point P.
Gradient of the tangent at P = lim(h→0) (f(x+h)-f(x))/h
This is often referred to as the gradient function written f'(x)
i.e f'(x) = lim(h→0) (f(x+h)-f(x))/h
This limit is used when calculating the derivative of f(x) from first principles. The derivative f'(x) is interpreted as the rate of change of y with respect to x. Hope you understood that [:)]
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Mr Worthington told us that in 1961!!!
And it hasn't changed.
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Mr Worthington told us that in 1961!!!
I presume that was your maths teacher.
Am I correct (with the calculus bit)?
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A legend!!