Applications of Implicit Differentiation

Section 13.4 Applications of Implicit Differentiation

In these examples, we will make use of the implicitdiff() and implicitplot() commands.

Example 13.2. Finding the Equation of a Tangent Line.

In this example, we will find the equations of the tangent lines to the circle of radius \(4\text{,}\) centred at the point \((1,1)\text{,}\) where \(x=3\text{.}\) Once these equations are assigned, we will plot the circle and the two tangent together in one single implicitplot() command.

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Aside

> circle := (x-1)\symbol{94}2 + (y-1)\symbol{94}2 = 16;

\begin{equation*} \displaystyle circle\, := \,{(x-1)}^{2}+{(y-1)}^{2}=16 \end{equation*}

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We need to find the \(y\)-coordinates by substituting \(x=3\) and solving for \(y\text{.}\)

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> subs(x=3, circle); yCoords:=solve(%);

\begin{equation*} \displaystyle 4+(y-1)^2 = 16 \end{equation*}

\begin{equation*} \displaystyle yCoords\, := \, 1+2\sqrt{3},\,1-2\sqrt{3} \end{equation*}

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The derivative \(\tfrac{dy}{dx}\) can be found using implicitdiff(). Then, by substituting the two points, we can find the slopes of both tangent lines.

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> dydx := implicitdiff(circle, y, x);

\begin{equation*} \displaystyle dydx\, := \,-{\frac {x-1}{y-1}} \end{equation*}

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Aside

> m1 := subs(x=3, y=yCoords[1], dydx);

\begin{equation*} \displaystyle m\mathit{1}\, := \,-\frac{1}{3}\sqrt{3} \end{equation*}

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> m2 := subs(x=3, y=yCoords[2], dydx);

\begin{equation*} \displaystyle m\mathit{2}\, := \,\frac{1}{3}\sqrt{3} \end{equation*}

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Recall that the equation of a line passing through \((x_0,y_0)\) with slope \(m\) can be written in the form

\begin{equation*} y = m (x - x_0) + y_0\text{.} \end{equation*}

If we wish to plot the circle and the two tangent lines together using a single implicitplot() command, they will need to be defined as equations involving an equals \(=\) sign. Each of these equations can then be expanded out to see the lines in their typical \(y=mx+b\) form.

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Aside

> line1 := y = m1*(x-3) + yCoords[1]; expand(%);

\begin{equation*} \displaystyle line\mathit{1}\, := \,y=-\frac{1}{3}\sqrt{3}(x-3)+1+2\sqrt{3} \end{equation*}

\begin{equation*} \displaystyle y=-\frac{1}{3}\sqrt{3}x+3\sqrt{3}+1 \end{equation*}

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> line2 := y = m2*(x-3) + yCoords[2]; expand(%);

\begin{equation*} \displaystyle line\mathit{2}\, := \,y=\frac{1}{3}\sqrt{3}(x-3)+1-2\sqrt{3} \end{equation*}

\begin{equation*} \displaystyle y=\frac{1}{3}\sqrt{3}x-3\sqrt{3}+1 \end{equation*}

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We can now plot the circle and the two lines together. Using the scaling=constrained parameter will produce a proper circle and avoid it being stretched either vertically or horizontally.

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Aside

> with(plots):
> implicitplot( [circle, line1, line2], x=-8..8, y=-8..8, 
    colour=[red,blue,green], scaling=constrained);

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Example 13.3. Orthogonal Curves.

In this example, we will show that the implicit curves \(x^2 - y^2 = 8\) and \(-xy =~3\) are always orthogonal (perpendicular) at their intersection points. We’ll start by assigning names to these two equations so that we can plot the two curves using implicitplot().

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> curve1 := x\symbol{94}2 - y\symbol{94}2 = 8;

\begin{equation*} \displaystyle curve\mathit{1}\, := \,{x}^{2}-{y}^{2}=8 \end{equation*}

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Aside

> curve2 := -x*y = 3;

\begin{equation*} \displaystyle curve\mathit{2}\, := \,-xy=3 \end{equation*}

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> with(plots):
> implicitplot([curve1, curve2], x=-5..5, y=-5..5, 
    colour=[red, blue], scaling=constrained);

Aside

From the graphs of these two curves, it appears that their intersections are perpendicular. This can be proven by showing that the derivative of one curve is equal to the negative reciprocal of the other, or that they multiply to equal \(-1\text{.}\)

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The intersection points are any points \((x,y)\) that satisfy both of the equations of these two implicit curves. They can be found using a single solve() (or fsolve()) command, so long as it is given both equations and the two variables, \(x\) and \(y\text{.}\)

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Aside

> solve( {curve1, curve2}, {x, y}, explicit = true);

\begin{equation*} \displaystyle \left\{ x=-3,y=1 \right\} ,\, \left\{ x=3,y=-1 \right\} ,\, \left\{ x=I,y=3\,I \right\} ,\, \left\{ x= -I,y=-3\,I \right\} \end{equation*}

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Points involving \(I\) are imaginary points and should not be considered. Therefore, the only two intersection points are \((-3,1)\) and \((3,-1)\text{,}\) which agrees with what we see from the graph.

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The derivatives of both curves can be found implicitly using the implicitdiff() command. They can be assigned two different names for easy comparison later.

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> dydx1 := implicitdiff(curve1, y, x);

\begin{equation*} \displaystyle dydx\mathit{1}\, := \,{\frac {x}{y}} \end{equation*}

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> dydx2 := implicitdiff(curve2, y, x);

\begin{equation*} \displaystyle dydx\mathit{2}\, := \,-{\frac {y}{x}} \end{equation*}

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To show that the slopes are negative reciprocals at a particular intersection point, the \(x\)- and \(y\)-values can be substituted into the two derivatives.

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> subs(x=-3, y=1, dydx1); subs(x=-3, y=1, dydx2);

\begin{equation*} \displaystyle -3 \end{equation*}

\begin{equation*} \displaystyle \frac{1}{3} \end{equation*}

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Alternatively, it can be shown that the derivatives are negative reciprocals of each other for any point \((x,y)\) where both slopes are defined.

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Aside

> dydx1*dydx2;

\begin{equation*} \displaystyle -1 \end{equation*}

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