Question

(a) The curve \(y = \frac{x}{{1 + {x^2}}}\) is called a serpentine. Find an equation of the tangent line to this curve at the point \(\left( {3,0.3} \right)\).

(b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

Short answer

(a) The equation of the tangent line at the point \(\left( {3,0.3} \right)\) is \(y = - 0.08x + 0.54\).

Step-by-step solution

The slope of the tangent

The slope of the tangent line to the curve is equal to the derivative of the function representing that curve at some given point.

The quotient rule of differentiation

If a function \(h\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\) where \(f\) and \(g\) are both differentiable, then the derivative of \(h\left( x \right)\) is as follows:

\(\begin{aligned}\frac{{dh}}{{dx}} & = \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)\\ & = \frac{{g\left( x \right) \cdot \frac{d}{{dx}}\left( {f\left( x \right)} \right) - f\left( x \right) \cdot \frac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\end{aligned}\)

The derivative of the given function

Differentiate \(f\left( x \right)\) with respect to \(x\) using quotient rule as follows:

\(\begin{aligned}f\left( x \right) & = \frac{x}{{1 + {x^2}}}\\f'\left( x \right) & = \frac{{\left( {1 + {x^2}} \right)\frac{d}{{dx}}\left( x \right) - x\frac{d}{{dx}}\left( {1 + {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\ & = \frac{{1 + {x^2} - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\ & = \frac{{1 - {x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}\end{aligned}\)

The slope of the tangent at \(\left( {3,0.3} \right)\)

Put \(x = 3\) in \(f'\left( x \right)\) and solve as follows:

\(\begin{aligned}f'\left( 3 \right) & = \frac{{1 - {{\left( 3 \right)}^2}}}{{{{\left( {1 + {{\left( 3 \right)}^2}} \right)}^2}}}\\ & = \frac{{1 - 9}}{{{{\left( {1 + 9} \right)}^2}}}\\ & = \frac{{ - 8}}{{100}}\\ & = - 0.08\end{aligned}\)

The equation of a line

The general equation of a line that passes through the point\(\left( {{x_1},{y_1}} \right)\)and has slope\(m\)is given below:

\(y - {y_1} = m\left( {x - {x_1}} \right)\)

The equation of a tangent line

Substitute the value of\({x_1}\),\({y_2}\)in the above formula to obtain the equation of tangent as follows:

\(\begin{aligned}y - 0.3 & = - 0.08\left( {x - 3} \right)\\y & = - 0.08x + 0.24 + 0.3\\y & = - 0.08x + 0.54\end{aligned}\)

Hence, the equation of the tangent line is \(y = - 0.08x + 0.54\).

Check the answer visually

The procedure to draw the graph of the above equation by using the graphing calculator is as follows:

To check the answer visually draw the graph of the function \({f_1}\left( x \right) = \frac{x}{{1 + {x^2}}}\)and \({f_2}\left( x \right) = 0.08x + 0.54\) by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\frac{x}{{1 + {x^2}}}\)in the\({Y_1}\)tab.
2. Select the “STAT PLOT” and enter the equation\( - 0.08x + 0.54\)in the\({Y_2}\)tab.
3. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function\({f_1}\left( x \right) = \frac{x}{{1 + {x^2}}}\)and \({f_2}\left( x \right) = - 0.08x + 0.54\) is shown below: