[Maths Class Notes] on Equation of Tangent and Normal Pdf for Exam

There are a number of applications of derivatives in science, engineering, and even social sciences. Some of the applications in the field of science are evaluating the rate of changes in quantities or finding equations of the tangent to a curve. We shall be discussing the concept of tangent and normal lines to a curve. 

Tangents and Normals

Tangents and normals are like any other straight lines at a given point on a curve. A tangent is parallel with the curve at the point, and the normal runs perpendicular to the curve. The equation of tangent and normal can be evaluated just like any other straight line. But to find the gradient of tangents and normals to a curve, students will need the derivative.

If the slope of the tangent to a curve y = f(x) at a point a is f'(a) (derivative of f(x) at point (a).

Then, equation of tangent in point-slope form is (y – f(a))/(x – a) = f'(a), and using equation mT × mN = −1, 

 

Equation of normal is 

 

(y – f(a))/(x – a) = −1/f'(a).

Learning about derivations leads to one of the conclusions as dy/dx geometrically representing at any point (x,y), the slope to the curve y =f(x). 

An important term to remember here is the gradient. A gradient of a curve at a point is defined as the slope of that tangent to the curve at that particular point. 

Moving further, the angle of the tangent plays an important role when finding the equation of the line. 

Therefore, below are some key points to remember about the angle of the tangent to the curve at a point:

1. If Ψ ( ≠ ⊼/2) is the angle at which the tangent at the curve point A (x,y) makes a positive direction of the x-axis, then in that case point A = tan Ψ = (dy/dx)A. The slope of the tangent to the curve remains y =f(x).

2. In this case, when the tangent to a curve is parallel to the x-axis, then Ψ = 0. This means that tan Ψ will be zero and therefore, (dy/dx)A will also be zero. 

3. When the tangent to the curve is parallel to the y-axis, then the Ψ = ⊼/2 = cot Ψ = 0 and therefore, (dy/dx)A becomes 0. 

4. The last case is when the tangent is parallel to the x-axis which means that Ψ ≠ 0. This further leads to the equation tan Ψ ≠ (dy/dx)A ≠ 0. Therefore, the slope of the normal to the curve at point A becomes A = −1/ (dy/dx)A.

Following these points above can help you progress further into finding the equation of tangent and normal.

Equation of Tangent 

The given curve is y =f(x) with point A (x1, y1). Therefore, the slope of the tangent becomes (dy/dx)x = x1 ; y = y1. 

Now since we have found how to calculate the slope, we can simply go forward with following the concept of coordinate geometry according to which an equation of a line is

(y − y1) = m(x − x1) , where m is the slope. 

We can now replace “m” from the equation above to the (dy/dx)x = x1 ; y = y1 . Now, the new equation of the tangent to the curve at point A (x1, y1) is

(y − y1) = (dy/dx)x = x1 ; y = y1 (x − x1).

The above-mentioned equation is the equation of the tangent formula. In summary, follow these three simple steps to find the equation of the tangent to the curve at point A (x1, y1).

Step 1: The first and foremost step should be finding (dy/dx) from the given equation of the curve y = f(x).

Step 2: The next step involves finding the value of (dy/dx) at point A (x1, y1). This method lets you calculate for “m” or the slope of that point. Therefore, m = (dy/dx)x = x1 ; y = y11.

Step 3: The last step is to remember the equation of the line from coordinate geometry which is (y − y1) = m(x − x1). 

The value of m is replaced with the slope of tangent equation that has been calculated above:

(y − y1) = (dy/dx)x = x1 ; y = y1 (x − x1).

Equation of Normal

When the equation of curve given is y = f(x), then finding the equation of the normal line at a curve point A (x1, y1) starts from finding the slope. The slope of the normal to the curve at point A (x1, y1) is A = A = −1/ (dy/dx)x = x1 ; y = y1.  

Since now we are aware of the slope, we can go forward with finding the equation of the normal line at the curve point  A (x1, y1), by using the formula derived from co-ordinate geometry :

(y − y1) = m(x − x1), where m is the slope. 

A point to note here is that although the equation for a normal line is the same as the tangent, the value of slope differs here. The slope for normal is −1/ (dy/dx)x = x1 ; y = y1 . 

(y − y1) = [ −1/ (dy/dx)x = x1 ; y = y1](x − x1)

Below are the three steps that can be followed one after the other when finding the equation of a normal line to a curve at point A (x1, y1):

Step 1: Determine the equation of the curve y = f(x). Once that is established, find (dy/dx) from the given equation of the curve; y = f(x).

Step 2: The next step is very important as it involves calculating the slope for the equation of the normal line to the curve at point A (x1, y1). To do this, find the value of  (dy/dx) at point A (x1, y1). The formula for the slope  of the normal is m =  −1/ (dy/dx)x = x₁ ; y = y₁ .

Step 3: The last step is replacing the value of slope in the equation (y − y1) = m(x − x1). After solving this, you have successfully found the equation for the normal.

The above-mentioned steps are simple to follow. When solving the equation of normal, refer to these steps in the order given. Following these steps can expedite the process of finding the equation of normal to the curve.

Have a quick revision of derivations in order to omit mistakes when attempting questions of this topic. Also, follow the simple steps mentioned above in order to gain a better understanding of the equation tangent and normal. 

Important Formulas of Tangent and Normal 

The following are the important formulas of tangent and normal to any curve at a given point.

• [frac{dy}{dx}|_{p}]  is the slope of the tangent to the curve y=f(x) at the point p.

• In a plane curve r=f(θ),

[tan phi = rfrac{dtheta}{dr}]

[(y – y_{1}) = frac{dy}{dx}|_{p}(x – x_{1})]

[(x – x_{1}) = frac{dy}{dx}|_{p} (y – y_{1}) ]

• Consider that a curve c is defined by y=f(x), and p is the length of the perpendicular from O(0,0) to the tangent at the point (x1,y1) of the curve. Then,

[p = frac{|y_{1} – x_{1}frac{dx}{dy}|}{sqrt{1-(frac{dy}{dx})^{2}}}]