Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2015 | June | Q#10
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Question
A curve with equation y=f(x) passes through the point (4,9).
Given that
, x > 0
a. find f(x), giving each term in its simplest form.
Point P lies on the curve.
The normal to the curve at P is parallel to the line 2y + x = 0
b. Find x coordinate of P.
Solution
a.
We are given;
We are given coordinates of a point on the curve (4,9).
We are required to find the equation of y in terms of x ie f(x).
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 the coordinates of point (4,9) in above equation;
Therefore, equation of the curve C is;
b.
We are given that point P lies on the curve and the normal to the curve at P is parallel to the line 2y + x = 0.
We are required to find the x coordinate of P.
We can find the x-coordinates of P through its gradient.
If we can find the gradient o the curve at point P, we can equate it with given expression of derivative (gradient) of the curve to find x-coordinates of point P.
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to
is:
Therefore, we need gradient of the curve at point P.
We are given that normal to the curve at point P has equation 2y + x = 0.
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 can find slope of the curve at point P if we can find slope of the normal to the curve at this point.
We are given equation of the normal to the curve at point P.
Slope-Intercept form of the equation of the line;
Where is the slopeof the line.
We can rearrange the given equation in slope-intercept form as follows.
Hence, slope of the normal to the curve at point P is .
Now we can find slope of the curve at point P.
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;
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