# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | May-Jun | (P1-9709/13) | Q#3

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Question

A curve is such that and passes through the point P(1,9). The gradient of the curve at P is 2.

i.
Find the value of the constant k.

ii.       Find the equation of the curve.

Solution

i.

We are given that; Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is: Hence, we have expression for the gradient of the curve.

We are also given that curve passes through the point P(1,9) and gradient of the curve at P is 2.

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

Gradient (slope) of the curve at a particular point can be found by substituting x- coordinates of that point in the expression for gradient of the curve; From the given information we can write;  Substituting ;    ii.

We can find equation of the curve from its derivative through integration;  We are given that; As we found in (i) that ; Therefore;    Rule for integration of is:  Rule for integration of is:      If a point lies on the curve , we can find out value of . We substitute values of and in the equation obtained from integration of the derivative of the curve i.e. .

We are also given that curve passes through the point P(1,9) and gradient of the curve at P is 2.

Substitution of x and y coordinates of point P in above equation;     Therefore equation of the curve is; 