Question

Show that the tangents to the curve \[y = 7{x^3} + 11\] at the points where \[x = 2\] and \[x = - 2\] are parallel?

Answer 1

Hint: Here, we are given a curve having two tangents and we need to find out if they are parallel or not. We know that two lines are said to be parallel when the slope of the 1st line is equal to the slope of the 2nd line. We will find the slope of the tangent by using \[\dfrac{{dy}}{{dx}}\] , then substitute the values of x and on solving this, we will get the final output.

Complete step by step answer:
Given a curve, \[y = 7{x^3} + 11\] and two tangents \[x = 2\]and \[x = - 2\]. First, we will find the slope of the tangent as below:
 \[y = 7{x^3} + 11\]
Differentiating this equation w.r.t. to x, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {7{x^3} + 11} \right)}}{{dx}}\]
As we know that, a derivative of a constant is zero and applying this, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 21{x^2} + 0\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 21{x^2}\] ------- (i)

Next, we need to show that the tangent at \[x = 2\]and \[x = - 2\] are parallel
\[ \Rightarrow \]Slope of tangent at \[x = 2\] = Slope of the tangent at \[x = - 2\]
So, for slope of the tangent to the curve at \[x = 2\] is \[{\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 2}}\]
Substituting the value of x in equation (i), we will get
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 2}} = 21{(2)^2}\]
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 2}} = 21(4)\]
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 2}} = 84\]

And, for slope of the tangent to the curve at \[x = - 2\] is \[{\left( {\dfrac{{dy}}{{dx}}} \right)_{x = - 2}}\]
Substituting the value of x in equation (i), we will get
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = - 2}} = 21{( - 2)^2}\]
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = - 2}} = 21(4)\]
\[ \therefore {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = - 2}} = 84\]
Thus, the slope of the tangent at \[x = 2\] is equal to the slope of the tangent at \[x = - 2\].So, they are parallel.

Hence, the tangents to the curve \[y = 7{x^3} + 11\] at the points where \[x = 2\]and \[x = - 2\] are parallel.

Note: In geometry, the tangent line (or tangent point) means a line or plane which intersects a curved line or surface at exactly one point. Derivatives can be defined as the varying rate of change of a function with respect to an independent variable. And, the process of finding the derivative is called differentiation. Also, the inverse process is called anti-differentiation.