The aim of this question is to deduce the equation of a tangent line of a curve at any point on the curve.
For any given function $ y = f(x) $, the equation of its tangent line is defined by the following equation:
\[ \boldsymbol{ y – y_1 = \frac{ dy }{ dx } ( x – x_1 ) } \]
Here, $ ( x_1 , y_1 ) $ is the point on the curve $ y = f(x) $ where the tangent line is to be evaluated and $ \dfrac{ dy }{ dx } $ is the value of the derivative of the subject curve evaluated at the required point.
Expert Answer
Given that:
\[ y = \sqrt{ x } \]
Calculating the derivative of $y$ with respect to $x$:
\[ \frac{ dy }{ dx } = \frac{ 1 }{ 2 \sqrt{ x } } \]
Evaluating above derivative at given point $( 81 , 9 )$:
\[ \frac{ dy }{ dx } |_{ ( 81 , 9 ) } = \frac{ 1 }{ 2 \sqrt{ 81 } } \]
\[ \frac{ dy }{ dx } |_{ ( 81 , 9 ) } = \frac{ 1 }{ 2 ( 9 ) } \]
\[ \frac{ dy }{ dx } |_{ ( 81 , 9 ) } = \frac{ 1 }{ 18 } \]
The equation of a tangent line with slope $\dfrac{ dy }{ dx }$ and point $( x_1 , y_1 )$ is defined as:
\[ y – y_1 = \frac{ dy }{ dx } ( x – x_1 ) \]
Substituting values of $ \dfrac{ dy }{ dx } = \dfrac{ 1 }{ 18 } $ and point $( x_1 , y_1 ) = ( 81 , 9 ) $ in above equation:
\[ y – 9 = \frac{ 1 }{ 18 } ( x – 81 ) \]
\[ y – 9 = \frac{ 1 }{ 18 } x – \frac{ 1 }{ 18 } 81 \]
\[ y – 9 = \frac{ 1 }{ 18 } x – \frac{ 9 }{ 2 } \]
\[ y = \frac{ 1 }{ 18 } x – \frac{ 9 }{ 2 } + 9 \]
\[ y = \frac{ 1 }{ 18 } x + \frac{ – 9 + ( 2 ) ( 9 ) }{ 2 } \]
\[ y = \frac{ 1 }{ 18 } x + \frac{ – 9 + 18 }{ 2 } \]
\[ \boldsymbol{ y = \frac{ 1 }{ 18 } x + \frac{ 9 }{ 2 } }\]
Numerical Result
\[ \boldsymbol{ y = \frac{ 1 }{ 18 } x + \frac{ 9 }{ 2 } }\]
Example
Find an equation of the tangent line to the curve $y = x$ at $(1, 10)$.
Here:
\[ \frac{ dy }{ dx } = 1 \]
Using the tangent equation with $ \dfrac{ dy }{ dx } = 1 $ and point $( x_1 , y_1 ) = ( 1 , 10 ) $:
\[ y – y_1 = \frac{ dy }{ dx } ( x – x_1 ) \]
\[ y – 10 = ( 1 ) ( x – 1 ) \]
\[ y = ( 1 ) ( x – 1 ) + 10 = x – 1 + 10 \]
\[ \boldsymbol{ y = x + 9 } \]