calculus
posted by Anonymous .
The slope of a curve is at the point (x,y) is 4x3. Find the curve if it is required to pass through the point (1,1).
Work...
4(1)3=1
y1=1(x1)
y=x

4x  3 is your slope, so it is the derivative.
then dy/dx = 4x + 3, integrate to get
y = 2x^2 + 3x + k
plug in (1,1) into that to get k, and you are done! 
why would I solve for the constant? I get k=4. What is meant by curve?

When they say "find the curve" they mean find the equation of function whose graph would be that curve.
your equation y = 2x^2 + 3x  4 would graph to be a parabola, and a parabola is a curve.
As to the constant, remember that if you differentiate an equation like
y = 2x^2 + 3x  4
you get y' = 4x + 3  0
so when we "antidifferentiate" that we really don't know what the value of the constant was, because its derivative would be zero no matter what the number was.
That is why we include a constant value of c or k to allow for that. Once we sub in the given point we then know the value of that constant
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