Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2012 | June | Q#7

 

Question

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

a.    

                    i.       Show that y is increasing when .

                  ii.       Solve the inequality .

b.   The curve C passes through the point P(2,3).

                    i.       Verify that the tangent to the curve at P is parallel to the x-axis.

                  ii.       The point Q(3,-1) also lies on the curve. The normal to the curve at Q and the tangent to  the curve at P intersect at the point R. Find the coordinates of R.

Solution

a.
 


i.
 

To test whether a function  is increasing or decreasing at a particular point , we  take derivative of a function at that point.

If  , the function  is increasing. 

If  , the function  is decreasing.

If  , the test is inconclusive.

 

Therefore, when y is increasing;

We are given that;

Therefore;


ii.
 

We are required to solve the inequality;

We solve the following equation to find critical values of ;

Now we have two options;

Hence the critical points on the curve for the given condition are 2 & .

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If  (‘a’ is positive) then parabola opens upwards  and its vertex is the minimum point on the graph.
If
 (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the  graph.

We recognize that  it is an upwards opening parabola.

Therefore conditions for  are;

b.
 


i.
 

If two lines are parallel to each other, then their slopes  and  are equal;

Therefore;

We know that , therefore so is 

Let’s now find slope of tangent to the curve at point P(2,3).

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

For the given case;

Therefore need to find gradient of the curve at point P(2,3).

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

We are given that;

We need gradient of the curve at point P(2,3).

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;

Hence;

Therefore, gradient of the curve at point P is ZERO, and hence so is the slope of the tangent to the  curve at this point.

It is evident that both x-axis and tangent to the given curve have same slopes and therefore are  parallel.


ii.
 

We are required to find the coordinates of point of intersection of tangent to the curve at point  P(2,3). and normal to the curve point Q(3,-1).

If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e.  coordinates of that point have same values on both lines (or on the line and the curve).  Therefore, we can equate  coordinates of both lines i.e. equate equations of both the lines (or the  line and the curve).

We need to find equations of both tangent and normal to find the coordinates of their point of  intersection.

Let’s find first equation of tangent to the curve at point P(2,3).

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 already have coordinates of a point on the tangent P(2,3). We have also found in (b:ii) slope of the tangent .

Now we can write equation of the tangent to the curve.

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

Now we find equation of the normal to the curve at point Q(3,-1).

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 already have coordinates of a point on the tangent Q(3,-1).

We need to find slope of the normal to the curve at point Q(3,-1).

If a line  is normal to the curve , then product of their slopes  and  at that point (where line  is normal to the curve) is;

Therefore, we need to find gradient of the curve at point Q(3,-1).

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

We are given that;

We need gradient of the curve at point Q(3,-1).

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;

Hence;

Therefore, gradient of the curve at point Q is;

Hence;

Now we can write equation of the normal to the curve.

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

 

Now we can find the coordinates of the point of intersection of tangent and normal.

Equation of the tangent is;

Equation of the normal is;

Equating both equations;

Single value of x indicates that there is only one intersection point.

With x-coordinate of point of intersection of two lines (or line and the curve) at hand, we can find the  y-coordinate of the point of intersection of two lines (or line and the curve) by substituting value  of x-coordinate of the point of intersection in any of the two equations. 

We choose;

Hence, coordinates of point of intersection of tangent and normal to the curve are (43,3).

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