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Question: How do you find the equation of the tangent and normal line to the curve \[y = {x^3}\] at \[x = 1\] ...

How do you find the equation of the tangent and normal line to the curve y=x3y = {x^3} at x=1x = 1 ?

Explanation

Solution

Hint : Here in this question, we have to find the equation of the tangent and normal line, given the x value then find the y value by substitute the x value in given equation, then differentiate the given equation with respect to xx and determine the slope of tangent at given point i.e., m=dydxm = \dfrac{{dy}}{{dx}} . Using slope of tangent determine the slope of normal i.e., m=−1m1m = \dfrac{{ - 1}}{{{m_1}}} ,then we have to find equation of tangent by using the equation of (y−y1)=m(x−x1)\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right) Where m is a slope of tangent and find equation of normal by using the equation of (y−y1)=m1(x−x1)\left( {y - {y_1}} \right) = {m_1}\left( {x - {x_1}} \right) Where m1 is a slope of normal.

Complete step-by-step answer :
The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point.
The normal line is the line that is perpendicular to the tangent line.
Consider the given equation of the curve
y=x3y = {x^3} ------(1)
Substitute xx as x1=1{x_1} = 1 in the equation (1) then
⇒y=13\Rightarrow y = {1^3}
⇒y=1\Rightarrow y = 1
Thus the point (x1,y1)\left( {{x_1},{y_1}} \right) = (1,1)\left( {1,1} \right)
Now find the Slope of tangent m at point (1,1)\left( {1,1} \right) is
Differentiate equation (1) w.r.t. xx .
⇒m=dydx=ddx(x3)\Rightarrow m = \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}({x^3})
⇒m=dydx=3x2\Rightarrow m = \dfrac{{dy}}{{dx}} = 3{x^2}
At (x1,y1)\left( {{x_1},{y_1}} \right) = (1,1)\left( {1,1} \right) , then
⇒m=(dydx)(1,1)=3(1)2\Rightarrow m = {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,1} \right)}} = 3{\left( 1 \right)^2}
⇒m=(dydx)(1,1)=3\Rightarrow m = {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,1} \right)}} = 3 -------(2)
Next find the Slope of normal m1 at point (1,1)\left( {1,1} \right) is
⇒m1=−1m=−1(dydx)(1,1)\Rightarrow {m_1} = \dfrac{{ - 1}}{m} = \dfrac{{ - 1}}{{{{\left( {\dfrac{{dy}}{{dx}}} \right)}_{(1,1)}}}}
⇒m1=−1m=−13\Rightarrow {m_1} = \dfrac{{ - 1}}{m} = \dfrac{{ - 1}}{3}
⇒m1=−13\Rightarrow {m_1} = - \dfrac{1}{3} --------(3)
Equation of tangent at point (x1,y1)\left( {{x_1},{y_1}} \right) = (1,1)\left( {1,1} \right) is
Consider the equation (y−y1)=m(x−x1)\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)
Substitute m from equation (2)
⇒(y−1)=3(x−1)\Rightarrow \left( {y - 1} \right) = 3\left( {x - 1} \right)
⇒y−1=3x−3\Rightarrow y - 1 = 3x - 3
Add 1 on both side
⇒y−1+1=3x−3+1\Rightarrow y - 1 + 1 = 3x - 3 + 1
On simplification, we get
⇒y=3x−2\Rightarrow y = 3x - 2
Equation of normal at point (x1,y1)\left( {{x_1},{y_1}} \right) = (1,1)\left( {1,1} \right) is
Consider the equation (y−y1)=m1(x−x1)\left( {y - {y_1}} \right) = {m_1}\left( {x - {x_1}} \right)
Substitute m1 by equation (3)
⇒(y−1)=−13(x−1)\Rightarrow \left( {y - 1} \right) = - \dfrac{1}{3}\left( {x - 1} \right)
⇒y−1=−13x+13\Rightarrow y - 1 = - \dfrac{1}{3}x + \dfrac{1}{3}
Add 1 on both side
⇒y−1+1=−13x+13+1\Rightarrow y - 1 + 1 = - \dfrac{1}{3}x + \dfrac{1}{3} + 1
On simplification, we get
⇒y=−13x+43\Rightarrow y = - \dfrac{1}{3}x + \dfrac{4}{3}
Hence, the equations of the tangent and the normal to the curve y=x3y = {x^3} at the point (x1,y1)\left( {{x_1},{y_1}} \right) = (1,1)\left( {1,1} \right) is
Equation of tangent is y=3x−2y = 3x - 2
Equation of normal is y=−13x+43y = - \dfrac{1}{3}x + \dfrac{4}{3} .
So, the correct answer is “ y=−13x+43y = - \dfrac{1}{3}x + \dfrac{4}{3} ”.

Note : The concept of the equation of tangent and normal comes under the concept of application of derivatives. Here the differentiation is applied on the equation of line or curve and then substitute the value of x. We should know about the general equation of a line. Hence these types of problems are solved by the above procedure.