1.8 - Finding a curve from its gradient

1.8 - Finding a curve from its gradient

A gradient function determines a family of curves; one point can select a particular member. You will find a curve's equation by integrating and using given information, recover a function from its second derivative, and check that every condition has been satisfied.

A point fixes the constant

Which curve with gradient 2x+42x+4 passes through (1,8)(1,8)? Integration gives y=x2+4x+Cy=x^2+4x+C, but the value of CC is still free. At the stated point, both x=1x=1 and y=8y=8 must hold, so 8=1+4+C8=1+4+C, giving C=3C=3.

This is how a point chooses one vertical position. In function notation, f(1)=8f(1)=8 means exactly the same thing as saying that (1,8)(1,8) lies on y=f(x)y=f(x).

Constant of integration

The constant of integration is the arbitrary constant left by reversing differentiation. Given a gradient on a connected interval, substituting one known point into the integrated function determines that constant.

Integrate first and then substitute. Substituting a point into the derivative instead would give the curve's gradient at that input, rather than its height.

Worked example 1: a polynomial gradient

Question. A curve passes through (2,7)(2,7) and has dy/dx=3x2−4dy/dx=3x^2-4. Find its equation.

Plan. Integrate the gradient, then use both coordinates of the point to find CC.

y=x3−4x+Cintegrate each term7=23−4(2)+Cuse the pointC=7.\begin{aligned} y&=x^3-4x+C &&\text{integrate each term}\\ 7&=2^3-4(2)+C &&\text{use the point}\\ C&=7. \end{aligned}

Answer. y=x3−4x+7y=x^3-4x+7.

Check. Its derivative is 3x2−43x^2-4, and its value at x=2x=2 is 8−8+7=78-8+7=7. Both conditions hold.

Keep the same gradient and the same input, but change the point's height. Only the integration constant changes.

Worked example 2: a different vertical position

Question. A curve passes through (2,10)(2,10) and has dy/dx=3x2−4dy/dx=3x^2-4. Find its equation.

Plan. Use the same antiderivative family and apply the new point.

y=x3−4x+C,10=8−8+C,C=10.y=x^3-4x+C,\qquad 10=8-8+C,\qquad C=10.

Answer. y=x3−4x+10y=x^3-4x+10.

Check. Differentiation again gives 3x2−43x^2-4. At x=2x=2, the value is 1010. This curve is 33 units above the curve in example 1, just as the two given heights differ by 33.

The argument also works for a root or reciprocal gradient, provided the given point belongs to its domain. If a gradient is specified on x>0x>0, do not extend the resulting answer across an excluded input without justification.

Watch out

From f(2)=7f(2)=7, write 7=F(2)+C7=F(2)+C after integrating. Writing 7=f′(2)7=f'(2) confuses a function's height with its gradient.

Use a fractional power to select a new curve.

Recovering gradient information

The second derivative describes how the gradient changes. To recover a curve from d2y/dx2d^2y/dx^2, reverse two differentiations in the correct order: first recover dy/dxdy/dx, then recover yy. Each integration introduces a constant.

For example, if d2y/dx2=6xd^2y/dx^2=6x, the gradient is dy/dx=3x2+Ady/dx=3x^2+A. A known gradient fixes AA. A second integration then gives y=x3+Ax+By=x^3+Ax+B, and a known point fixes BB. The first constant becomes the coefficient of xx; it does not disappear.

You already know

The second derivative is the derivative of the first derivative. Reversing the process one step at a time keeps the two unknown constants attached to the information that determines them.

Use different letters for the constants so their jobs remain clear.

Worked example 3: returning from a second derivative

Question. A curve has d2y/dx2=6xd^2y/dx^2=6x. Its gradient at x=1x=1 is 55, and it passes through (1,4)(1,4). Find its equation.

Plan. Integrate once and use the gradient, then integrate again and use the point.

dydx=3x2+A5=3+Agradient at x=1A=2y=x3+2x+B4=1+2+Bpoint (1,4)B=1.\begin{aligned} \frac{dy}{dx}&=3x^2+A\\ 5&=3+A &&\text{gradient at }x=1\\ A&=2\\ y&=x^3+2x+B\\ 4&=1+2+B &&\text{point }(1,4)\\ B&=1. \end{aligned}

Answer. y=x3+2x+1y=x^3+2x+1.

Check. Differentiating twice gives 6x6x; the first derivative at 11 is 3+2=53+2=5; the function value there is 1+2+1=41+2+1=4.

Change only the specified gradient. This alters the coefficient of xx, so the second constant must also adjust to keep the same point.

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