There is a big difference between someone explicitly telling you that you need a haircut, and someone implicitly telling you. An implicit comment might be something like, "I think you look better with short hair" while an explicit one would be, "Go cut your hair!". Equations in calculus have a similar difference between explicit and implicit. So read on to figure out how to manage implicit relations!
Implicit Relations in Calculus
The types of equations you are probably used to working with in calculus are equations of the form
\[ y=\text{expression in terms of }x.\]
These types of equations define explicit relations, in other words, one where you can explicitly solve for \(y\). However, many of the interesting equations you will run into in calculus are not quite so simple.
Sometimes, they are of the form:
\[ \text{expression in terms of }x \text{ and }y=\text{expression in terms of }x \text{ and }y.\]
These equations define implicit relations.
An implicit relation in mathematics is one where you cannot explicitly solve for one variable to write the relation as a function.
Let's look at an example.
In the equation, \(x^3+y^3=6xy\) can you explicitly solve for \(y\)?
Solution:
No, there is no way to write this as \(y=\text{expression in terms of }x. \) So, it is an implicit relation. It is a particularly famous one, in fact, called the folium of Descartes. If you graph the relation, it looks like the graph below.
Fig. 1 - Graph of the folium of Descartes.
Notice that this is not a function since it fails the vertical line test!
The vertical line test says: "If a vertical line cuts a graph twice this graph is not an explicit function". For a reminder about the vertical line test and explicitly defined functions, see the article Functions.
Implicit Relations in Math
So what is the general form of an implicit relation? It is an equation of the form
\[f(x,y)=g(x,y)\]
where \(f(x,y)\) and \(g(x,y)\) are functions of two variables.
In the context of calculus, an implicit relation is defined by an equation where the dependent variable is not isolated on one side of the equation. While you can extend this to more than two variables, for now, let's stick with two since they can often be graphed nicely.
A circle centered at the origin with radius \(2\) is an example of implicit relation given by the equation:
\[x^2+y^2=4.\]
To write it in the form given above, let \(f(x,y) = x^2+y^2\) and \(g(x,y) = 4.\) No one said there had to be an actual \(x\) or \(y\) in both \(f\) and \(g.\)
Even though these equations are not functions, you can still study them.
A binary relation (on the real numbers) is a set of ordered pairs \((x,y)\) of real numbers.
A binary relation could be a function! It might or might not be able to be written as an equation.
The equation \(y-x^2=0\) defines the binary relation
\[\{(x,x^2)\text{ for }x\in\mathbb{R}\}.\]
Notice that this is in fact, a function since it can be written as \(y=x^2\).
Writing the equation as a binary relation treats \(y-x^2=0\) as a rule that tells you whether a particular pair of numbers should be in its corresponding relation.
Pick any pair of real numbers \((a,b)\). If \(b-a^2=0\), then it belongs to the relation corresponding to the equation \(y-x^2=0\). This only happens when \(b=a^2\). Otherwise, \((a,b)\) does not belong to the relation.
Not all binary relations are functions!
Look at the relation defined by the equation \(x^2+y^2=1\). It is a circle centered at the origin of radius \(1\). Writing it as a binary relation, you get
\[\left\{ \left( x,\pm\sqrt{1-x^2}\right)\text{ for }x\in[-1,1]\right\}.\]
So this binary relation is not a function.
In fact, \(x^2+y^2=1\) is written implicitly, while you can write this explicitly as \(y=\pm\sqrt{1-x^2} \).
Not all relations, and not even all functions, can be defined by an equation. In fact, most functions in math cannot be defined by an equation! If you reached into a hat that contained all possible functions on the real numbers, odds are you would pick one that cannot be defined by an equation or algorithm. We call such functions incomputable functions.
You will need some terminology to talk about binary relations. Given any binary relation and any element \((a,b)\) in that relation:
\(a\) is an input of that relation;
\(b\) is an output of that relation; and
a function is a particular type of relation that can have only one output for every input.
In other words, if \((a,b)\) is an element of a function and \((a,c)\) is also an element of a function, then \(b=c\).
Graphs of Implicit Relations
The graph of an implicit relation is the set of points in the plane that correspond to ordered pairs in that relation. Graphs and relations are not quite the same thing. A relation is just a set of ordered pairs of numbers. The graph of a relation is a geometric interpretation of that relation; it assigns ordered pairs of numbers to points in the plane.
The relation corresponding to the equation \(y^2=x^3-x+0.2\) is the set
\[\left\{ ( x,y): \; y^2=x^3-x+0.2 \right\}\]
The graph corresponding to this relation looks like this:
Fig. 2 - The graph corresponding to a relation is a way of geometrically viewing the elements of that relation.
This is an example of an elliptic curve, a type of curve that is important in Number Theory. These types of curves were essential to Andrew Wiles' proof of Fermat's Last Theorem, and are also significant in cryptography.
Notice that this curve is not a function since it fails the vertical line test!
Let's look at another example.
This is another elliptic curve, \(y^2 = x^3 - x + 1\). Notice that it is also not a function, but it is an implicit relation.
Fig. 3 - Another example of an elliptic curve.
If you want to know when you can take an implicit relation and break it into pieces which are actually functions, you will need the Implicit Function Theorem. This involves taking partial derivatives.
For a reminder on partial derivatives, see Implicit Differentiation.
Derivatives of Implicit Relations
The derivative of an implicit relation can be found using partial derivatives. Let's take a quick look at an example, and for more information on how to take partial derivatives, see the article Implicit Differentiation.
Find the derivative of the implicit relation defined by the equation \(y^2=x^3-x+1\).
Solution:
Begin by differentiating both sides of the relation:
\[\dfrac{\mathrm{d}}{\mathrm{d}x}(y^2)=\dfrac{\mathrm{d}}{\mathrm{d}x }(x^3-x+1).\]
The right-hand side of the equation can be differentiated as usual:
\[\begin{align}\frac{\mathrm{d}}{\mathrm{d}x }(x^3-x+1)&=\dfrac{\mathrm{d}}{\mathrm{d}x }(x^3)+\dfrac{\mathrm{d}}{\mathrm{d}x }(-x)+\dfrac{\mathrm{d}}{\mathrm{d}x }(1) \\ &= 3x^2 -1+0 \\ &=3x^2-1\end{align}\]
Remember that \(y\) is a function of \(x\). So taking the derivative of the left side of the equation you get:
\[\dfrac{\mathrm{d}}{\mathrm{d}x }(y^2)=2yy'\]
where you have used the Chain Rule.
Putting those together, you get
\[ 2yy' =3x^2-1 ,\]
so
\[y' = \frac{3x^2-1 }{2y}.\]
You may wonder about tangent lines for implicit relations. For more information and examples of that see the article Finding Tangent Lines Implicitly.
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