Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2013 | January | Q#6

 

Question

The gradient, , of a curve at the point (x,y) is given by

The curve passes through the point P(1,4).

a.   Find the equation of the tangent to the curve at the point P, giving your answer in the form .

b.   Find the equation of the curve.

Solution

a.
 

We are required to find the equation of the tangent to the curve at the point P.

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).

We are given coordinates of appoint on the tangent P(1,4). However, we need to find the slope of  tangent.

The slope of a curve  at a particular point is equal to the slope of the tangent to the curve at the  same point;

Therefore, we now first find gradient of the curve at point P(1,4).

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is:

We are already given;

However, we need gradient of the curve at point P(1,4).

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;

Therefore, we substitute x-coordinate of point  P(1,4) in expression of gradient of the curve.

Therefore gradient of the curve at point P(1,4);

Hence;

Now we can write equation of the tangent to the curve at point P(1,4).

Point-Slope form of the equation of the line is;

b.
 

We are required to find the equation of the curve when we are given coordinates of a point P(1,4)  on the curve and;

We can find equation of the curve from its derivative through integration; 

Therefore;

Rule for integration of  is:

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. .

Therefore, substituting coordinates of point P(1,4);

Hence, equation of the curve is;

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