Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2018 | May-Jun | (P2-9709/22) | Q#5

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Question

A curve has equation

Find the equation of the tangent to the curve at the point where it crosses the y-axis.

Solution

We are given that curve with equation  and we are required to find the equation of  the tangent to the curve at point where it crosses the y-axis i.e.  .

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 have coordinates of point on the tangent (and curve) as (0,y).

Therefore, first we need to find the y-coordinate of the curve.

Corresponding value of y coordinate can be found by substituting value of x in equation of the  curve.

Hence coordinates of the point on the curve where we are required to find equation of the tangent  are

Next, we need slope of the tangent 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 find gradient (slope) of the curve at point where tangent  intersects the curve,  then we can find slope of the tangent.

Hence, we need gradient of the curve at point  .

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 are given that;

Therefore;

Rule for differentiation of  is:

First we  differentiate  using Product Rule.

If  and  are functions of , and if , then;

If , then;

Let and , then;

To find from an implicit equation, differentiate each term with respect to , using the chain rule to  differentiate any function  as .

Rule for differentiation of  is:

To differentiate  we utilize Chain Rule.

If we define , then derivative of  is;

If we have  and then derivative of  is;

Let ;

Rule for differentiation of  is;

Since ;

Rule for differentiation of  is:

Rule for differentiation of  is:

Hence;

Next we differentiate .

To find from an implicit equation, differentiate each term with respect to , using the chain rule to  differentiate any function  as .

Finally, we differentiate .

Rule for differentiation of  is:

Now we can finally write;

We are looking for gradient of the curve at point .

Therefore;

Therefore slope of the curve at point . is;

Hence, slope of the tangent to the curve at this point is;

Now we can write equation of the tangent as follows.

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