Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

Implicit Differentiation

At this point, we want to consider curves that do not correspond to a function. By the vertical line test, such curves can be circles, ellipses or cardiods (see Exercise 27 on page 189 of the textbook). These curves are given by equations; for example the equation ${\left(x-k\right)}^2+{\left(y-l\right)}^2=r^2$ corresponds to a circle with centre at $\left(k,l\right)$ and radius $r$. Such curves have tangent lines at some points, and we want to find the slopes of these tangent lines. But the slopes of tangent lines are given by the derivative, and differentiation is defined only for functions. One way to overcome this problem would be to decompose the curve into parts or pieces such that each piece corresponds to a function, and then differentiate.

Observe, on page 182 of the textbook, how by solving for $y$ in the equation $x^2+y^2=25$, we can decompose the circle into an upper and a lower half-circle, each represented by a function:

$f\left(x\right)=\sqrt{25-x^2}\text{and }g\left(x\right)=-\sqrt{25-x^2}\text{.}$

Then,

$x^2+{\left(f\left(x\right)\right)}^2=25\text{ and }{x}^2+{\left(g\left(x\right)\right)}^2=25\text{.}$

We can differentiate these functions:

${\displaystyle f^{\prime }\left(x\right)=-{\displaystyle \frac{x}{\sqrt{25-x^2}}}\text{ and }{g}^{\prime }\left(x\right)={\displaystyle \frac{x}{\sqrt{25-x^2}}}\text{.}}$

The slope of the tangent line at the point $\left(3,4\right)$ in the upper half-circle is

${\displaystyle f^{\prime }\left(3\right)=-{\displaystyle \frac{3}{4}}\text{;}}$

and the tangent at the point $\left(3,-4\right)$ in the lower half-circle is

${\displaystyle g^{\prime }\left(3\right)={\displaystyle \frac{3}{4}}\text{.}}$

We can see that $f^{\prime }\left(5\right)$ and $f^{\prime }\left(-5\right)$ are undefined; that is, the tangent lines at $\left(5,0\right)$ and $\left(-5,0\right)$ are vertical.

Figures 2 and 3 on page 183 of the textbook show the graph of the curve $x^3+y^3=6xy$, and its decomposition into different pieces, each of which corresponds to a function.

The problem here is that, explicitly to find the functions that correspond to each piece, we would have to solve for $y$ in the equation $x^3+y^3=6xy$, and this is not an easy task. However, we can assume that it can be done, since we have the pieces that show that there are functions corresponding to these curves.

So we say that these functions are implicit in the equation; that is, we can interpret the equation to mean that $y=f\left(x\right)$ for some function $f\left(x\right)$, and $x^3+{\left(f\left(x\right)\right)}^3=6x\left(f\left(x\right)\right)$.

To differentiate with respect to $x$, we can apply the differentiation rules, so that

$3x^2+3{\left(f\left(x\right)\right)}^2{f}^{\prime }\left(x\right)=6x\left(f^{\prime }\left(x\right)\right)+6f\left(x\right)=6\left(x{f}^{\prime }\left(x\right)+f\left(x\right)\right)\text{.}$

Since $y=f\left(x\right)$ and $y^{\prime }=f^{\prime }\left(x\right)$, we conclude that

$3x^2+3y^2{y}^{\prime }=6\left(x{y}^{\prime }+y\right)\text{,}$

and we can solve for $y^{\prime }$:

$3y^2{y}^{\prime }-6x{y}^{\prime }=6y-3x^2\text{ and }{y}^{\prime }\left(3y^2-6x\right)=3\left(2y-x^2\right)\text{.}$

Therefore,

${\displaystyle y^{\prime }={\displaystyle \frac{2y-x^2}{y^2-2x}}\text{.}}$

The point $\left(3,3\right)$ is on the curve, because $3^3+3^3=6\left(3\right)\left(3\right)$, and for $x=3$, $y=3$, the derivative at $\left(3,3\right)$ is

${\displaystyle y^{\prime }={\displaystyle \frac{2\left(3\right)-3^2}{3^2-2\left(3\right)}}=-1\text{,}}$

so the slope of the tangent line at the point $\left(3,3\right)$ is $-1$.

In the same manner, you can verify that

• the point ${\displaystyle \left({\displaystyle \frac{8}{3}},{\displaystyle \frac{4}{3}}\right)}$ is on the curve, and
• for ${\displaystyle x={\displaystyle \frac{8}{3}}\text{ and }y={\displaystyle \frac{4}{3}}\text{,}}$

  ${\displaystyle y^{\prime }={\displaystyle \frac{2\left({\displaystyle {\displaystyle \frac{4}{3}}}\right)-{\left({\displaystyle {\displaystyle \frac{8}{3}}}\right)}^2}{{\left({\displaystyle {\displaystyle \frac{4}{3}}}\right)}^2-2\left({\displaystyle {\displaystyle \frac{8}{3}}}\right)}}={\displaystyle \frac{5}{4}}\text{.}}$

The slope of the tangent line at the point ${\displaystyle \left({\displaystyle \frac{8}{3}},{\displaystyle \frac{4}{3}}\right)\text{ is }{\displaystyle \frac{5}{4}}\text{.}}$

So, by implicitly differentiating the equation (by assuming that $y=f\left(x\right)$), we solved the problem without having to find explicitly the function that corresponds to the curve that passes through the designated point.

Example 4.58. An astroid is shown in Exercise 28 on page 187 of the textbook; its equation is $x^{2∕3}+y^{2∕3}=4\text{.}$

As before, if we assume that $y=f\left(x\right)$, then the equation is

$x^{2∕3}+{\left(f\left(x\right)\right)}^{2∕3}=4\text{,}$

and by the differentiation rules,

${\displaystyle {\displaystyle \frac{2}{3}}{x}^{-1∕3}+{\displaystyle \frac{2}{3}}{\left(f\left(x\right)\right)}^{-1∕3}{f}^{\prime }\left(x\right)=0\text{.}}$

Replacing $y=f\left(x\right)$ and $y^{\prime }=f^{\prime }\left(x\right)$, we obtain

${\displaystyle {\displaystyle \frac{2}{3}}{x}^{-1∕3}+{\displaystyle \frac{2}{3}}{y}^{-1∕3}{y}^{\prime }=0\text{,}}$

and solving for $y^{\prime }$ yields

${\displaystyle y^{\prime }=-{\displaystyle \frac{y^{1∕3}}{x^{1∕3}}}=-\sqrt[3]{{\displaystyle \frac{y}{x}}}\text{.}}$

The points $\left(0,8\right)$ and $\left(8,0\right)$ are on the curve. For $x=0$ and $y=8$, we know that $y^{\prime }$ is undefined, because there is no tangent line at this point. For $x=8$ and $y=0\text{,}$ we know that $y^{\prime }=0\text{,}$ so the tangent line is horizontal. Examine the curve and convince yourself that this is indeed the case.

Example 4.59. If we are careful to remember that $y=f\left(x\right)\text{,}$ we may not need to replace $y$ by $f\left(x\right)$ in an equation, and we may be able to differentiate without mistakes.

In the equation $x^4+x{y}^4=x sin y\text{,}$  we differentiate, obtaining

$4x^3+y^4+4x{y}^3{y}^{\prime }=sin y+x{y}^{\prime }cos y\text{.}$

Solving for $y^{\prime }$ yields

${\displaystyle y^{\prime }={\displaystyle \frac{sin y-4x^3-y^4}{4x{y}^3-x cos y}}\text{.}}$

Families of similar curves are represented by equations that differ only by a constant. For example, circles with centre at the origin $\left(0,0\right)\text{,}$ and radii $1$ and $2$ have equations $x^2+y^2=1$ and $x^2+y^2=4\text{,}$ respectively. In general, the family of circles with centre at the origin and different radii $r$ is represented by the equation $x^2+y^2=r^2$ for any positive real number $r\text{.}$

Example 4.60. To show that the curves represented by the equations $3x^2+2y^2=5$ and $y^3=x^2$ are orthogonal, we must first find their points of intersection by solving the two equations simultaneously.

Substituting $x^2=y^3$ into the other equation, we obtain $3y^3+2y^2=5$; hence, $y^2\left(3y+2\right)=5\text{.}$ Since $5=3+2\text{,}$ we find that $y=1$ is a solution of this equation (there might be others, but we do not need them at this time).

So, if $y=1$ then $x^2=y^3=1$; and therefore, $x=1$ and $x=-1\text{.}$ The points of intersection are $\left(1,1\right)$ and $\left(-1,1\right)\text{.}$

Now we have to see whether the slopes of the tangent lines at these points are perpendicular. Implicitly differentiating both equations and solving for $y^{\prime }\text{:}$

${\displaystyle 6x+4y{y}^{\prime }=0\text{ and }{y}^{\prime }=-{\displaystyle \frac{6x}{4y}}=-{\displaystyle \frac{3x}{2y}}\text{.}}$(1)

${\displaystyle 3y^2{y}^{\prime }=2x\text{ and }{y}^{\prime }={\displaystyle \frac{2x}{3y^2}}\text{.}}$(2)

For $x=1$ and $y=1\text{,}$ the slope of the tangent line of the curve in ($1$) is $y^{\prime }=-3∕2\text{,}$ and the slope of the tangent line of the curve in ($2$) is $y^{\prime }=2∕3\text{.}$ Therefore, the tangent lines are perpendicular at $\left(1,1\right)\text{.}$

For $x=-1$ and $y=1\text{,}$ we have $y^{\prime }=3∕2$ in ($1$) and $y^{\prime }=-2∕3$ in ($2$).

Therefore, the tangent lines are also perpendicular at $\left(-1,1\right)\text{,}$ and the curves are orthogonal.

Example 4.61. To see if the families of parabolas $x=a{y}^2$ and ellipses $2x^2+y^2=2b$ have orthogonal trajectories, we need to see if the slopes of the tangent lines are perpendicular. Implicitly differentiating both equations and solving for $y^{\prime }$ yields

${\displaystyle 1=2ay{y}^{\prime }\text{ and }{y}^{\prime }={\displaystyle \frac{1}{2ay}}\text{.}}$(1)

${\displaystyle 4x+2y{y}^{\prime }=0\text{ and }{y}^{\prime }=-{\displaystyle \frac{2x}{y}}\text{.}}$(2)

Now observe that from the equation $x=a{y}^2=\left(ay\right)y\text{,}$ we have $ay=x∕y\text{;}$ hence, in ($1\text{),}$ we have $y^{\prime }=y∕\left(2x\right)\text{.}$

The slopes of the tangent lines of the curves are $y^{\prime }=y∕\left(2x\right)$ and $y^{\prime }=-2x∕y\text{;}$ hence, the tangent lines are perpendicular, and the curves have orthogonal trajectories.