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Calculus of Powers

Review question

Given three clues, can we find the equation of the curve?

Ref: R7734

The gradient of a curve is ax+b at all points, where a and b are constants. Find the equation of the curve, given that it passes through the points (0,4) and (1,3), and that the tangent at (1,3) is parallel to the line y=x.

Let’s say the equation of the curve is y=f(x), so y=ax+b. This means that y=ax+bdx=ax22+bx+c for some constant of integration, c. (This is a consequence of the fundamental theorem of calculus.)

As the curve passes through the point (0,4) we know that a×022+b×0+c=4c=4, and as it passes through the point (1,3), we also have that a×122+b×1+c=3a2+b+4=3a+2b=2. Now the tangent to the curve at (1,3) is parallel to the line y=x, which has gradient 1, so a+b=4 (since y=ax+b).

We therefore have simultaneous equations in a and b, which are a+2b=2,a+b=1.

Subtracting, we have b=3, and hence a=4. The equation of the curve is therefore y=2x23x+4.