# 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

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;