**Tangents and Normal to Curves**

For any curve, is the gradient function. At any point on the curve, at that point, gives the gradient of the tangent at the point. The derivation of y with respect to x at x = x_{1} is denoted.

Table of Contents

x = x_{1}

Recall that the equation of the line of gradient m through (x_{1}, y_{1}) is y – y_{1} = m(x – x_{1}).

From this equation, we can easily obtain the equation of the tangent.

The straight line, perpendicular to the tangent at the point of contact of the tangent to the curve is called the **Normal** to the curve.

If m^{1} is the gradient of the Normal, and m is the gradient of the tangent at the point of contact of the tangent to the curve, then

m^{1} =

So at the point (x_{1}, y_{1}) the equation of the normal is y – y_{1} = (x – x_{1})

Find the equation of the tangent and the normal to the curve y = 2x^{3} – x^{2} + 3x + 1 at the point x = 1.

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**Solution**

Given that y = 2x^{3} – x^{2} + 3x + 1

6x^{2} – 2x + 3

x = x_{1}= 6 – 2 + 3 = 7

If m is the gradient of the tangent at x = 1, then m = 7.

At the point x = 1; y = 2 – 1 + 3 + 1 = 5.

The equation of the tangent at the point x = 1 is

y – 5 = 1(x – 1)

y – 5 7x– 2

If m^{1} is the gradient of the normal at x = 1, then

m^{1} =

Hence, the equation of the normal at x = 1, is y – 5 = (x – 1)

7(y – 5) = -1 (x – 1)

7y – 35 = -x + 1

7y + x – 36 = 0

Find the equation of the tangent to the curve x^{2}y + y^{2}x + 3x – 13 = 0 at the point (1, 2)

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**Solution**

x^{2}y + y^{2}x + 3x – 13 = 0

x^{2} + 2xy + 3y^{2}x y^{2} + 3 = 0

(x^{2} + 3y^{2}x) + 2xy + y^{3} + 3 = 0

(x^{2} + 3y^{2}x) = -2xy – y^{3} – 3

-2xy – y^{3} -3

x^{2} + 3y^{2}x

If m is the gradient of the tangent at the point (1, 2) then

m = x = 1, y = 2

= =

The equation of the tangent is therefore

y – 2 = (x – 1)

13 (y – 2) = -15 (x – 1)

13y – 26 = -15x + 15

13y + 15x – 41 = 0

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**A curve is defined by**

f(x) = x^{3 }– 6x^{3} – 15x – 1. Find

(i) the derivation f(x) with respect to x;

(ii) the gradient of the curve at the point where x = 1;

(iii) the maximum and the minimum points.

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**Find the maximum and minimum points of the curve **y = x^{3}– x^{2} – 5x and sketch the curve.

**A curve passes through the point (1, 0) and its gradient at any point p(x, y) is 3x ^{2} – 1.**

(i) Find the equation of the curve;

(ii) Sketch the curve, indicating all turning points and the point of intersection with the axes.

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Sketch the curve y = x^{3} – 6x^{2} + 9x.

Find:

(i)the equation of the tangent to the curve y = x^{3} + x2 – 8x + 2 at the point A(1, -4);

(ii) the coordinate of the point where the tangent meet the x – axis.

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**Evaluation **

**Find the equation of the normal to each of the following curve**

(4) y = (2x – 3) (x + 2) at x = 1

**General Evaluation**

**Find the equation of the tangent to each of the following curve at the given points**

(1) y = x^{2} – 3x – 4 at x = 1

(2) y = x^{3} + 2x^{2} – 3x + 1 at x = -1

(3) y = 1 – 2x + 5x^{2} – x^{3} at x = 3

Find the equation of the normal to each of the following curve

(5) y = 6 – 2x + 3x^{2} – 2x^{3} at x = 0

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**Weekend Assignment**

A curve y = 4x^{3} – 2x^{2} + 7x + 5 at point x = 3, find the

(1) Gradient of its tangent

(a) 12 (b) 108 (c) 103

(d) 115

(2) Gradient of its normal

(a)

(3) Equation of the tangent

(a)y = 108x – 116 (b) y = 103x – 193 (c) y = 115x – 90

(d) y = 12x – 309

(4) Equation of the normal

(a) 103y + x – 11 = 0 (b) 103x + y – 116 = 0 (c) 116x + x – 119 = 0

(d) 116x + y – 103 = 0

(5) Find the gradient of this curve y = 2x^{2} – 5x + 8 at point x = 1

(a) 1 (b) -1 (c) 2 (d) 3

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**Theory**

(1) Find the gradient of the tangent and the normal of the curve y = x^{3} – 6x^{2} – 15x – 1 at point x = 1

(2) A curve passes through the point (1, 0) and its gradient at any point p(x, y) is 3x^{2} – 1, find the equation of the curve.

See also