Implicit Differentiation: Learn It 2

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Implicit Differentiation: Learn It 2

Finding Tangent Lines Implicitly

We can now apply implicit differentiation to effectively find equations of tangent lines for curves defined by more complex equations. This method not only broadens our understanding of calculus applications but also reinforces the analytical power of implicit differentiation.

Find the equation of the line tangent to the curve $x^2+y^2=25$ at the point $(3,-4)$ .

Although we could find this equation without using implicit differentiation, using that method makes it much easier. In an earlier example, we found $\frac{dy}{dx}=-\frac{x}{y}$ .

The slope of the tangent line is found by substituting $(3,-4)$ into this expression. Consequently, the slope of the tangent line is

$\frac{dy}{dx}|_{(3,-4)} =-\frac{3}{-4}=\frac{3}{4}$ .

Using the point $(3,-4)$ and the slope $\frac{3}{4}$ in the point-slope equation of the line, we then solve for $y$ to obtain the equation

$y=\frac{3}{4}x-\frac{25}{4}$ .

The circle with radius 5 and center at the origin is graphed. A tangent line is drawn through the point (3, −4). — Figure 2. The line $y=\frac{3}{4}x-\frac{25}{4}$ is tangent to $x^2+y^2=25$ at the point $(3,−4)$ .

Find the equation of the line tangent to the graph of $y^3+x^3-3xy=0$ at the point $\left(\frac{3}{2},\frac{3}{2}\right)$ (Figure 3). 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. Finding the tangent line to the folium of Descartes at $(\frac{3}{2},\frac{3}{2})$ .

Begin by finding $\frac{dy}{dx}.$

\begin{matrix} \frac{d}{d x} (y^{3} + x^{3} - 3 x y) = \frac{d}{d x} (0) 3 y^{2} \frac{d y}{d x} + 3 x^{2} - (3 y + 3 x \frac{d y}{d x}) = 0 \frac{d y}{d x} = \frac{3 y - 3 x^{2}}{3 y^{2} - 3 x} . \end{matrix}

$\begin{array}{l} \frac{d}{dx}(y^3+x^3-3xy) = \frac{d}{dx}(0) \\ \\ 3y^2\frac{dy}{dx}+3x^2-(3y+3x\frac{dy}{dx}) = 0 \\ \\ \frac{dy}{dx} = \frac{3y-3x^2}{3y^2-3x}. \\ \\ \end{array}$

Next, substitute $(\frac{3}{2},\frac{3}{2})$ into $\frac{dy}{dx}=\frac{3y-3x^2}{3y^2-3x}$ to find the slope of the tangent line:

\frac{d y}{d x} |_{(\frac{3}{2}, \frac{3}{2})} = - 1

$\frac{dy}{dx}|_{(\frac{3}{2},\frac{3}{2})}=-1$

Finally, substitute into the point-slope equation of the line and solve for $y$ to obtain

y = - x + 3

$y=−x+3$

In a simple video game, a rocket travels in an elliptical orbit whose path is described by the equation $4x^2+25y^2=100$ . 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 $(0,0)$ . If the rocket fires a missile when it is located at $\left(3,\frac{8}{5}\right)$ , where will it intersect the $x$ -axis?

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

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

Differentiating, we have

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

$8x+50y\frac{dy}{dx}=0$

Solving for $\frac{dy}{dx}$ , we have

\frac{d y}{d x} = - \frac{4 x}{25 y}

$\dfrac{dy}{dx}=-\dfrac{4x}{25y}$

The slope of the tangent line is $\frac{dy}{dx}|_{(3,\frac{8}{5})}=-\frac{3}{10}$ . 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 $(\frac{25}{3},0)$ .

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “3.8 Implicit Differentiation” here (opens in new window).