The applications of derivatives are:
determining the rate of change of quantities finding the equations of tangent and normal to a curve at a point finding turning points on the graph of a function which in turn will help us to locate points at which largest or smallest value (locally) of a function occurs. In this article, we will learn to use differentiation to find the equation of the tangent line and the normal line to a curve at a given point.
Tangent and Normal Equation We know that the equation of the straight line that passes through the point (x0 , y0 ) with finite slope “m” is given as
y – y0 = m (x – x0 )
It is noted that the slope of the tangent line to the curve f(x)=y at the point (x0 , y0 ) is given by
\(\begin{array}{l}\frac{dy}{dx}]_{(x_{0}, y_{0})} (=f'(x_{0}))\end{array} \)
Therefore, the equation of the tangent (x0 , y0 ) to the curve y=f(x) is
y – y 0 = f ′(x 0 )(x – x 0 )
Also, we know that normal is the perpendicular to the tangent line. Hence, the slope of the normal to the curve f(x)=y at the point (x0 , y0 ) is given by -1/f’(x0 ), if f’(x0 ) ≠ 0.
Hence, the equation of the normal to the curve y=f(x) at the point (x0 , y0 ) is given as:
y-y0 = [-1/f’(x0 )] (x-x0 )
The above expression can also be written as
(y-y 0 ) f’(x 0 ) + (x-x 0 ) = 0
Points to Remember If a tangent line to the curve y = f (x) makes an angle θ with x-axis in the positive direction, then dy/dx = slope of the tangent = tan = θ. If the slope of the tangent line is zero, then tan θ = 0 and so θ = 0 which means the tangent line is parallel to the x-axis. In this case, the equation of the tangent at the point (x0 , y0 ) is given by y = y0 If θ →π/2, then tan θ → ∞, which means the tangent line is perpendicular to the x-axis, i.e., parallel to the y-axis. In this case, the equation of the tangent at (x0 , y0 ) is given by x = x0 📌 Related Posts:
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Equation of Tangent and Normal Problems Go through the below tangent and normal problems:
Example 1:
Find the equation of a tangent to the curve y = (x-7)/[(x-2)(x-3)] at the point where it cuts the x-axis.
Solution:
As the point cut at the x-axis, then y=0. Hence, the equation of the curve, if y=0, then the value of x is 7. (i.e., x=7). Hence, the curve cuts the x-axis at (7,0)
Now, differentiate the equation of the curve with respect to x, we get
dy/dx = [(1-y)(2x-5)] / [(x-2)((x-3)]
dy/dx](7, 0) = (1-0)/[(5)(4)] = 1/20
Hence, the slope of the tangent line at (7, 0) is 1/20.
Therefore, the equation of the tangent at (7, 0) is
Y-0 = (1/20)(x-7)
20y-x+7 = 0.
Example 2:
Find the equation of tangent and normal to the curve x(⅔) + y(⅔) = 2 at (1, 1)
Solution:
Given curve: x(⅔) + y(⅔) = 2
Finding Equation of Tangent:
Now, differentiate the curve with respect to x, we get
(⅔)x(-⅓) + (⅔)y(-⅓) dy/dx = 0
The above equation can be written as:
dy/dx = -[y/x]⅓
Hence, the slope of the tangent at the point (1, 1) is dy/dx](1,1) = -1
Now, substituting the slope value in the tangent equation, we get
Equation of tangent at (1, 1) is
y-1 = -1(x-1)
y+x-2 = 0
Thus, the equation of tangent to the curve at (1, 1) is y+x-2 =0
Finding Equation of Normal:
The slope of the normal at the point (1, 1) is
= -1/slope of the tangent at (1, 1)
= -1/ -1
=1
Therefore, the slope of the normal is 1.
Hence, the equation of the normal is
y-1 = 1(x-1)
y-x = 0
Therefore, the equation of the normal to the curve at (1, 1) is y-x =0
Practice problems Solve the following problems:
Calculate the slope of the tangent to the curve y=x3 -x at x=2. Determine the slope of the tangent to the curve y=x3 -3x+2 at the point whose x-coordinate is 3. Find the equation of tangent and normal to the curve y = x3 at (1, 1). Find the equation of normal at the point (am2 , am3 ) for the curve ay2 =x3 . The slope of the normal to the curve y=2x2 + 3 sin x at x=0 is (a)3 (b) -3 (c)⅓ (d) -⅓