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Differentiation Question

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Dibss:
Practice and you'll get used to it eventually. (:

gradient of line y=3x-2 is 3 b/c in y=mx+c form m is the gradient.
So now we have to find the point on the curve where gradient is 3.
dy/dx= 4x-1
tangent to curve parallel where 4x-1=3
4x=4
x=1
at x=1, y=2-1-1=0
therefore coordinates should be (1,0)
(:

Arthur Bon Zavi:
Differentiation is easy and Integration is easier.

Tohru Kyo Sohma:

--- Quote from: Dibss (Whisper) on December 31, 2010, 06:59:27 pm ---Practice and you'll get used to it eventually. (:

gradient of line y=3x-2 is 3 b/c in y=mx+c form m is the gradient.
So now we have to find the point on the curve where gradient is 3.
dy/dx= 4x-1
tangent to curve parallel where 4x-1=3
4x=4
x=1
at x=1, y=2-1-1=0
therefore coordinates should be (1,0)
(:
thank you so much wispher
i got it now...........;)

--- End quote ---

Tohru Kyo Sohma:
i have a doubt here;
The production cost per kilogram C(in thousand pounds) when x kilogram of a chemical are made, is given by;
C= 3x+100/x , x>0. find the value for x for which the cost is minimum and the minimum cost??

elemis:
Differentiate the equation. Set dy/dx = 0 and solve for x.

Determine if the stationary point is a minimum or maximum. Select the minimum point and plug it into the original equation.

This is the minimum cost.

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