Question

Find the equation of a curve that passes through the point (1;4) and dy/dx=4x^3+4x

Accepted Answer

We have dy/dx = 4x^3 + 4x.
dy/dx = 4x^3 + 4x
=> dy = (4x^3 + 4x) dx
Integrate both the sides
Int [ dy ] = Int [ (4x^3 + 4x) dx ]
=> y = 4x^4 / 4 + 4*x^2 / 2
=> y = x^4 + 2*x^2 + C
As the curve passes through (1 , 4)
4 = 1^4 + 2*1^2 + C
=> 4 = 1 + 2 + C
=> C = 1
This gives y = x^4 + 2*x^2 + 1
The required equation of the curve is y = x^4 + 2*x^2 + 1

Answer

To determine the equation of the curve, we'll have to determine the antiderivative of the given expression.
Int dy = Int (4x^3+4x)dx
We'll use the property of integrals to be additive:
Int (4x^3+4x)dx = Int 4x^3 dx + Int 4xdx
Int (4x^3+4x)dx = 4 Int x^3dx + 4Int xdx
Int (4x^3+4x)dx = 4*x^4/4 + 4*x^2/2 + C
We'll simplify and we'll get:
Int (4x^3+4x)dx = x^4 + 2x^2 + C
The expression represents a family of curves that depends on the values of the constant C.
We know, from enunciation that the point  (1 , 4) is located on the curve. Therefore it's coordinates will verify the equation of the curve.
4 = (1)^4 +2*(1)^2 + C
4 = 1 + 2 + C
4 = 3 + C
C = 4 - 3
C = 1
The equation of the curve, whose derivative is dy/dx=4x^3+4x , is the complete square: y = x^4 + 2x^2 + 1 = (x^2 + 1)^2.

Calculus Questions and Answers

Find the equation of a curve that passes through the point (1;4) and dy/dx=4x^3+4x

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