Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#6

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Question

The curve C has equation

 , x>0

a.  Find  in its simplest form.

b.  Find an equation of the tangent to C at the point where x=2.

Solution

a.
 

We are given;

We are required to find .

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

Therefore;

Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:

b.
 

We are required to find the equation of tangent to the curve C at point where x=2.

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 x-coordinate of a point on the tangent ie x=2.

We can find y-coordinate of this point on the tangent by substituting x=2 in equation of the curve  found in (a), because the same point also lies on the curve (this is the only point where tangent and  curve intersect).

Therefore, we substitute x=2 in the following equation of curve C;

Hence, coordinates of a point on the tangent to curve C are (2,-15).

We need to find slope of tangent at in order to write its equation.

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, if we can find slope of the curve C at point P(2,-15) then we can find slope of the tangent  to the curve at this point.

We need to find the gradient of the curve C at point P(2,-15).

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;

We have found in (a);

For gradient of the curve at point P(2,-15), substitute  in derivative of the equation of the curve. 

Therefore;

With coordinates of a point on the tangent P(2,-15) and its slope  in hand, we can write equation of the tangent.

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

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