Finding Tangent Lines Implicitly

Now that we have seen the technique of implicit differentiation, we can apply it to the problem of finding equations of tangent lines to curves described by equations.

Example 3.71

Finding a Tangent Line to a Circle

Find the equation of the line tangent to the curve \(x^{2} + y^{2} = 25\) at the point \(\left(\right. 3 , −4 \left.\right) .\)

Solution

Although we could find this equation without using implicit differentiation, using that method makes it much easier. In Example 3.68, we found \(\frac{d y}{d x} = - \frac{x}{y} .\)

The slope of the tangent line is found by substituting \(\left(\right. 3 , −4 \left.\right)\) into this expression. Consequently, the slope of the tangent line is \(\begin{aligned} \frac{d y}{d x} \left|\right. \\ _{\left(\right. 3 , −4 \left.\right)} = - \frac{3}{−4} = \frac{3}{4} . \end{aligned}\)

Using the point \(\left(\right. 3 , −4 \left.\right)\) and the slope \(\frac{3}{4}\) in the point-slope equation of the line, we obtain the equation \(y = \frac{3}{4} x - \frac{25}{4}\) (Figure 3.31).

The circle with radius 5 and center at the origin is graphed. A tangent line is drawn through the point (3, −4).

Figure 3.31 The line \(y = \frac{3}{4} x - \frac{25}{4}\) is tangent to \(x^{2} + y^{2} = 25\) at the point (3, −4).

Example 3.72

Finding the Equation of the Tangent Line to a Curve

Find the equation of the line tangent to the graph of \(y^{3} + x^{3} - 3 x y = 0\) at the point \(\left(\right. \frac{3}{2} , \frac{3}{2} \left.\right)\) (Figure 3.32). This curve is known as the folium (or leaf) of Descartes.

A folium is shown, which is a line that creates a loop that crosses over itself. In this graph, it crosses over itself at (0, 0). Its tangent line from (3/2, 3/2) is shown.

Figure 3.32 Finding the tangent line to the folium of Descartes at \(\left(\right. \frac{3}{2} , \frac{3}{2} \left.\right) .\)

Solution

Begin by finding \(\frac{d y}{d x} .\)

\[\begin{aligned} \frac{d}{d x} \left(\right. y^{3} + x^{3} - 3 x y \left.\right) & = & \frac{d}{d x} \left(\right. 0 \left.\right) \\ 3 y^{2} \frac{d y}{d x} + 3 x^{2} - \left(\right. 3 y + \frac{d y}{d x} 3 x \left.\right) & = & 0 \\ \frac{d y}{d x} & = & \frac{3 y - 3 x^{2}}{3 y^{2} - 3 x} . \end{aligned}\]

Next, substitute \(\left(\right. \frac{3}{2} , \frac{3}{2} \left.\right)\) into \(\frac{d y}{d x} = \frac{3 y - 3 x^{2}}{3 y^{2} - 3 x}\) to find the slope of the tangent line:

\[\begin{aligned} \frac{d y}{d x} \left|\right. \\ _{\left(\right. \frac{3}{2} , \frac{3}{2} \left.\right)} = −1 . \end{aligned}\]

Finally, substitute into the point-slope equation of the line to obtain

\[y = − x + 3 .\]

Example 3.73

Applying Implicit Differentiation

In a simple video game, a rocket travels in an elliptical orbit whose path is described by the equation \(4 x^{2} + 25 y^{2} = 100 .\) The rocket can fire missiles along lines tangent to its path. The object of the game is to destroy an incoming asteroid traveling along the positive x-axis toward \(\left(\right. 0 , 0 \left.\right) .\) If the rocket fires a missile when it is located at \(\left(\right. 3 , \frac{8}{5} \left.\right) ,\) where will it intersect the x-axis?

Solution

To solve this problem, we must determine where the line tangent to the graph of

\(4 x^{2} + 25 y^{2} = 100\) at \(\left(\right. 3 , \frac{8}{5} \left.\right)\) intersects the x-axis. Begin by finding \(\frac{d y}{d x}\) implicitly.

Differentiating, we have

\[8 x + 50 y \frac{d y}{d x} = 0 .\]

Solving for \(\frac{d y}{d x} ,\) we have

\[\frac{d y}{d x} = - \frac{4 x}{25 y} .\]

The slope of the tangent line is \(\frac{d y}{d x} \left|\right. _{\left(\right. 3 , \frac{8}{5} \left.\right)} = - \frac{3}{10} .\) The equation of the tangent line is \(y = - \frac{3}{10} x + \frac{5}{2} .\) To determine where the line intersects the x-axis, solve \(0 = - \frac{3}{10} x + \frac{5}{2} .\) The solution is \(x = \frac{25}{3} .\) The missile intersects the x-axis at the point \(\left(\right. \frac{25}{3} , 0 \left.\right) .\)

Checkpoint 3.49

Find the equation of the line tangent to the hyperbola \(x^{2} - y^{2} = 16\) at the point \(\left(\right. 5 , 3 \left.\right) .\)

This lesson is part of:

Derivatives

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