SUNY Geneseo Department of Mathematics
Wednesday, October 14
Math 221 02
Fall 2020
Prof. Doug Baldwin
What are “explicit” and “implicit” in this reading, and what does dy/dx mean in connection with them?
“Implicit” and “explicit” refer to ways of writing equations that relate 2 variables.
In an “explicit” equation, one variable appears alone, usually on the left side of the “=,” while the other side gives a formula for computing that variable from the other. There is a clear sense that the variable on the right side of the equation is the independent variable, i.e., the one you can plug values in to, while the variable on the left is the dependent variable, i.e., the one whose value is controlled by the other side; information “flows” from the right side of the equation to the left. This is the kind of equation you are probably most used to seeing.
An implicit equation, on the other hand, simply states some relationship involving both variables. How either variable depends on the other is implicit in that relationship. There’s no sense of independent or dependent variables, or of information about one variable’s value “flowing” to the other. For example, you may have seen the equation for a unit circle as a relationship involving x2 and y2; this is a good example of an implicit relationship between x and y:
Regardless of how the relationship between x and y is defined, the notation dy/dx refers to how fast y changes as x changes.
Rate of change makes sense for both explicit and implicit equations. For example, the circle defined implicitly above has a clear sense of tangents to the circle and their slopes, just as explicitly defined curves have.
Thus both implicit and explicit functions have derivatives; implicit differentiation is the technique for finding them when given an implicit relationship.
The next SI session is Sunday, 5:00 - 6:30. A Zoom link will be available about 1 hour before.
From “Implicit Differentiation” in section 3.8 of the textbook, and the implicit differentiation discussion.
3x2 - 5y2 = 1
t2 - 2ty + y = 0
cos( z2 + x2 ) = 1
The first 2 of these have good solutions in the discussion, but we can certainly talk about them if you want.
The last example doesn’t have anything in the discussion yet, so let’s start with it.
All implicit differentiations have two main steps: first, differentiate both sides of the equation, and second isolate dy/dx on one side of the resulting equation (see the discussion for a slightly longer version of this sequence of steps). Start with the first of these steps:
Notice that we used the chain rule twice here: first to differentiate the cos(z2+x2), but then a second time to differentiate z2. This second use is because z is implicitly a function of x, the variable we’re differentiating with respect to, so any time we take a derivative involving z we use the chain rule, treating z as the inner function, and so multiplying by the derivative of z. This use of the chain rule is key to making implicit differentiation work.
After differentiating both sides of the equation, we went on to isolate dz/dx. This just relied on algebra, namely moving terms without dz/dx to one side of the equation, and then dividing by whatever was multiplied by dz/dx to leave dz/dx by itself:
Try the first example from the discussion, 3x2 - 5y2 = 1.
Once again, start by differentiating both sides of the equation:
Then isolate dy/dx:
Continue implicit differentiation, examining more complicated contexts, further implications, etc.
Please read “Finding Tangent Lines Implicitly” in section 3.8 of the textbook, and keep following and contributing to the implicit differentiation discussion, by class time Thursday.