Past Papers’ Solutions  Assessment & Qualification Alliance (AQA)  AS & A level  Mathematics 6360  Pure Core 1 (6360MPC1)  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 xaxis.
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 xaxis 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 (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope 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.
PointSlope 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 (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope 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.
PointSlope 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 xcoordinate of point of intersection of two lines (or line and the curve) at hand, we can find the ycoordinate of the point of intersection of two lines (or line and the curve) by substituting value of xcoordinate 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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