@@ -54,7 +48,7 @@
- Additional Practice Exercises
+ Calculus Practice
- Foundational Exercises
+ Foundations
- Additional Practice Exercises
+ Calculus Practice
- Foundational Exercises
+ Foundations
This section is about limits, the behavior of a function's output value as the input value is getting closer
and closer (but not equal to) a specified number. A function can be described by a graph, a table of values,
@@ -110,7 +110,7 @@
- Additional Practice Exercises
+ Calculus PracticeLimits from a graph
diff --git a/source/sec-1-8-tan-line-approx.xml b/source/sec-1-8-tan-line-approx.xml
index 11860d3..b57a176 100644
--- a/source/sec-1-8-tan-line-approx.xml
+++ b/source/sec-1-8-tan-line-approx.xml
@@ -118,7 +118,7 @@
- Additional Practice
+ Calculus Practice
diff --git a/source/sec-2-1-elem-rules.xml b/source/sec-2-1-elem-rules.xml
index d7d7b89..eac1d89 100644
--- a/source/sec-2-1-elem-rules.xml
+++ b/source/sec-2-1-elem-rules.xml
@@ -137,7 +137,7 @@
- Additional Practice Exercises
+ Calculus Practice
diff --git a/source/sec-2-2-sin-cos.xml b/source/sec-2-2-sin-cos.xml
index 419fa65..2b808c3 100644
--- a/source/sec-2-2-sin-cos.xml
+++ b/source/sec-2-2-sin-cos.xml
@@ -69,7 +69,7 @@
- Additional Practice Exercises
+ Calculus Practice
diff --git a/source/sec-2-3-prod-quot.xml b/source/sec-2-3-prod-quot.xml
index 1b56938..c263bd7 100644
--- a/source/sec-2-3-prod-quot.xml
+++ b/source/sec-2-3-prod-quot.xml
@@ -34,7 +34,7 @@
- Foundational Exercises
+ Foundations
Coming soon.
@@ -42,7 +42,7 @@
- Additional Practice Exercises
+ Calculus Practice
diff --git a/source/sec-2-4-other-trig.xml b/source/sec-2-4-other-trig.xml
index 2214eb7..8b40f82 100644
--- a/source/sec-2-4-other-trig.xml
+++ b/source/sec-2-4-other-trig.xml
@@ -30,7 +30,7 @@
- Additional Practice
+ Calculus Practice
diff --git a/source/sec-2-5-chain.xml b/source/sec-2-5-chain.xml
index c6cbfc0..5821b99 100644
--- a/source/sec-2-5-chain.xml
+++ b/source/sec-2-5-chain.xml
@@ -40,7 +40,7 @@
- Foundational Exercises
+ Foundations
One of the most useful ways of constructing functions from other functions is composition: doing one
function and then doing another function to the result. Rather than substituting a number in for the input
@@ -112,7 +112,7 @@
- What does it mean to say that a curve is an implicit function of x,
- rather than an explicit function of x?
+ Calculate \frac{dy}{dx} when y depends on x but isn't
+ an explicit function of x.
-
- How does implicit differentiation enable us to find a formula for
- \frac{dy}{dx} when y is an implicit function of x?
-
-
-
-
-
- In the context of an implicit curve,
- how can we use \frac{dy}{dx} to answer important questions about the tangent line to the curve?
+ Use \frac{dy}{dx} to answer questions about the slope (or equation) of the tangent line to the curve
+ even when y is not an explicit function of x.
-
- Introduction
-
- In all of our studies with derivatives so far,
- we have worked with functions whose formula is given explicitly in terms of x.
- But there are many interesting curves whose equations involving x and y are impossible to solve for y in terms of x.
-
-
-
-
At left, the circle given by x^2 + y^2 = 16. In the middle, the portion of the circle x^2 + y^2 = 16 that has been highlighted in the box at left. And at right, the curve given by x^3 - y^3 = 6xy.
-
-
-
-
- Perhaps the simplest and most natural of all such curves are circles.
- Because of the circle's symmetry,
- for each x value strictly between the endpoints of the horizontal diameter,
- there are two corresponding y-values.
- For instance, in Figure,
- we have labeled A = (-3,\sqrt{7}) and B = (-3,-\sqrt{7}),
- and these points demonstrate that the circle fails the vertical line test.
- Hence, it is impossible to represent the circle through a single function of the form y = f(x).
- But
- portions of the circle can be represented explicitly as a function of x,
- such as the highlighted arc that is magnified in the center of Figure.
- Moreover, it is evident that the circle is locally linear,
- so we ought to be able to find a tangent line to the curve at every point.
- Thus, it makes sense to wonder if we can compute
- \frac{dy}{dx} at any point on the circle,
- even though we cannot write y explicitly as a function of x.
-
-
-
- We say that the equation x^2 + y^2 = 16 defines yimplicitly
- as a function of x.
- The graph of the equation can be broken into pieces where each piece can be defined by an explicit function of x.
- For the circle,
- we could choose to take the top half as one explicit function of x,
- namely y = \sqrt{16 - x^2} and the bottom half as the explicit function y = -\sqrt{16 - x^2}.
- The equation for the circle defines an
- implicit function of x.
- implicit function
-
-
-
- The righthand curve in Figure
- is called the folium of Descartes
- folium of Descartes
- and is just one of many fascinating possibilities for implicitly given curves.
-
-
-
- How can we find an equation for
- \frac{dy}{dx} without an explicit formula for y in terms of x?
- To begin answering this question, the following preview activity reminds us of some ways we can compute derivatives of functions in settings where the function's formula is not known.
-
-
-
-
-
-
-
-
- Implicit Differentiation
-
- We begin our exploration of implicit differentiation with the example of the circle given by x^2 + y^2 = 16.
- How can we find a formula for \frac{dy}{dx}?
-
-
-
- By viewing y as an implicit
- function of x,
- we think of y as some function whose formula f(x) is unknown,
- but which we can differentiate.
- Just as y represents an unknown formula,
- so too its derivative with respect to x,
- \frac{dy}{dx}, will be
- (at least temporarily)
- unknown.
-
-
-
- So we view y as an unknown differentiable function of x and
- differentiate both sides of the equation with respect to x.
-
- \frac{d}{dx} \left[ x^2 + y^2 \right] = \frac{d}{dx} \left[ 16 \right]
- .
-
-
-
- On the right,
- the derivative of the constant 16 is 0,
- and on the left we can apply the sum rule,
- so it follows that
-
- \frac{d}{dx} \left[ x^2 \right] + \frac{d}{dx} \left[ y^2 \right] = 0
- .
-
-
-
- Note carefully the different roles being played by x and y.
- Because x is the independent variable,
- \frac{d}{dx} \left[x^2\right] = 2x.
- But y is the dependent variable and y is an implicit function of x.
- Recall Preview Activity,
- where we computed \frac{d}{dx}[f(x)^2].
- Computing \frac{d}{dx}[y^2] is the same,
- and requires the chain rule,
- by which we find that \frac{d}{dx}[y^2] = 2y^1 \frac{dy}{dx}.
- We now have that
-
- 2x + 2y \frac{dy}{dx} = 0
- .
-
-
-
- We solve this equation for \frac{dy}{dx} by
- subtracting 2x from both sides and dividing by 2y.
-
- \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}
- .
-
-
-
-
- Let's think further about the result that \frac{dy}{dx} = -\frac{x}{y}.
- First, notice that this expression for the derivative involves both x and y.
- This makes sense because
- there are two corresponding points on the circle for each value of x between -4 and 4,
- and the slope of the tangent line is different at each of these points.
- Second, this formula is entirely consistent with our understanding of circles.
- The slope of the radius from the origin to the point (a,b) is m_r = \frac{b}{a}.
- The tangent line to the circle at (a,b) is perpendicular to the radius,
- and thus has slope m_t = -\frac{a}{b},
- as shown in Figure.
- In particular,
- the slope of the tangent line is zero at (0,4) and (0,-4),
- and is undefined at (-4,0) and (4,0).
-
-
-
-
The circle given by x^2 + y^2 = 16 with point (a,b) on the circle and the tangent line at that point, with labeled slopes of the radial line, m_r, and tangent line, m_t.
-
-
-
-
-
- All of these observations about the circle are consistent with the formula \frac{dy}{dx} = -\frac{x}{y}.
-
-
-
-
-
- For the curve given implicitly by x^3 + y^2 - 2xy = 2,
- find the slope of the tangent line at (-1,1).
-
-
-
-
- We begin by differentiating the curve's equation implicitly.
- Taking the derivative of each side with respect to x,
-
- \frac{d}{dx}\left[ x^3 + y^2 - 2xy \right] = \frac{d}{dx} \left[ 2 \right]
- ,
- by the sum rule and the fact that the derivative of a constant is zero, we have
-
- \frac{d}{dx}[x^3] + \frac{d}{dx}[y^2] - \frac{d}{dx}[2xy] = 0
- .
-
-
-
- For the three derivatives we now must execute,
- the first uses the simple power rule,
- the second requires the chain rule
- (since y is an implicit function of x),
- and the third necessitates the product rule
- (again since y is a function of x).
- Applying these rules, we now find that
-
- 3x^2 + 2y\frac{dy}{dx} - [2x \frac{dy}{dx} + 2y] = 0
- .
-
-
+
+
+
- We want to solve this equation for \frac{dy}{dx}.
- To do so, we first collect all of the terms involving
- \frac{dy}{dx} on one side of the equation.
-
- 2y\frac{dy}{dx} - 2x \frac{dy}{dx}= 2y - 3x^2
- .
+ Recognize explicit and implicit functions of x, in both algebraic and graphical forms.
-
+
+
- Then we factor the left side to isolate \frac{dy}{dx}.
-
- \frac{dy}{dx}(2y - 2x) = 2y - 3x^2
- .
+ Solve an equation for \frac{dy}{dx}, including factoring out \frac{dy}{dx} from multiple terms.
-
+
+
- Finally, we divide both sides by (2y - 2x) and conclude that
-
- \frac{dy}{dx} = \frac{2y-3x^2}{2y-2x}
- .
+ The slope of a horizontal line is 0, and the slope of a vertical line is undefined.
+
+
+
+ Foundations
+
+ What are implicit and explicit functions?
- Note that the expression for
- \frac{dy}{dx} depends on both x and y.
- To find the slope of the tangent line at (-1,1),
- we substitute the coordinates into the formula for \frac{dy}{dx},
- using the notation
-
- \left. \frac{dy}{dx} \right|_{(-1,1)} = \frac{2(1)-3(-1)^2}{2(1)-2(-1)} = -\frac14
- .
+ You may remember learning that a function has to pass the Vertical Line Test which is a visual way
+ of seeing if each input resulted in exactly 1 output, not 2 or 3 or more outputs for the same input. We now would call that an explicit
+ function of x; an implicit function does not need to pass the Vertical Line Test. It would
+ be more accurate to say that y depends on x, but it is not necessarily a function of x.
+
+
+
+
+
+ Slopes of horizontal and vertical lines
- If we plotIn Desmos, entering the equation x^3 + y^2 - 2xy = 2 will automatically graph the corresponding curve. the implicit curve and its tangent line, as shown in Figure,
- we see that the slope we have found matches the graph and confirms the local linearity of this curve at (-1,1).
+ In implicit differentiation problems, the formula for
+ \frac{dy}{dx} is usually a fraction. To find where the tangent line is horizontal, set the
+ numerator equal to 0 and solve for either x or y in terms of the other variable, then
+ substitute that expression into the original equation for the curve to solve the equation of one variable.
+ To find where the tangent line is vertical, we do the same thing but with setting the denominator
+ of \frac{dy}{dx} equal to 0.
-
-
The curve x^3 + y^2 - 2xy = 2 and its tangent line at (-1,1).
-
-
-
-
-
-
- Example
- shows that it is possible when differentiating implicitly to have multiple terms involving \frac{dy}{dx}.
- We use addition and subtraction to collect all terms involving
- \frac{dy}{dx} on one side of the equation,
- then factor to get a single term of \frac{dy}{dx}.
- Finally, we divide to solve for \frac{dy}{dx}.
-
-
-
- We use the notation
-
- \left. \frac{dy}{dx} \right|_{(a,b)}
-
- to denote the evaluation of
- \frac{dy}{dx} at the point (a,b).
- This is analogous to writing f'(a) when f' depends on a single variable.
-
-
-
- There is a big difference between writing
- \frac{d}{dx} and \frac{dy}{dx}.
- For example,
-
- \frac{d}{dx}[x^2 + y^2]
-
- gives an instruction to take the derivative with respect to x of the quantity x^2 + y^2,
- presumably where y is a function of x.
- On the other hand,
-
- \frac{dy}{dx}(x^2 + y^2)
-
- means the product of the derivative of y with respect to x with the quantity x^2 + y^2.
- Understanding this notational subtlety is essential.
-
-
-
-
-
- It is natural to ask where the tangent line to a curve is vertical or horizontal.
- The slope of a horizontal tangent line must be zero,
- while the slope of a vertical tangent line is undefined.
- Often the formula for
- \frac{dy}{dx} is expressed as a quotient of functions of x and y, say
-
- \frac{dy}{dx} = \frac{p(x,y)}{q(x,y)}
- .
-
-
-
- The tangent line is horizontal precisely when the numerator is zero and the denominator is nonzero,
- making the slope of the tangent line zero.
- If we can solve the equation
- p(x,y) = 0 for either x and y in terms of the other,
- we can substitute that expression into the original equation for the curve.
- This gives an equation in a single variable,
- and if we can solve that equation we can find the point(s) on the curve where p(x,y) = 0.
- At those points, the tangent line is horizontal.
-
-
-
- Similarly, the tangent line is vertical whenever
- q(x,y) = 0 and p(x,y) \ne 0,
- making the slope undefined.
-
-
-
-
-
-
-
-
- Summary
-
-
-
+ Slopes of horizontal and vertical lines
+
+
+ Note that one of the lines below is horizontal and one line is vertical. Which is which?
+
+
+
+
+ Slopes of horizontal and vertical lines
+
+
+ Note that one of the lines below is horizontal and one line is vertical. Which is which?
+
+
+
+
+
+
+ Solving an equation with multiple variables
+
+ In implicit differentiation problems, you will often end up with an equation involving x, y,
+ and \frac{dy}{x}, and you will need to solve the equation for what \frac{dy}{dx} equals.
+
+
+ It would be good to have some practice solving equations for a variable, including factoring out by the variable.
+
+
+
+ Previous derivative rules and notation
+
+ Implicitly defined functions often involve sums, products, quotients, and compositions. This means that
+ you need to recognize the structure of the different terms in the function and apply the appropriate
+ derivative rules correctly.
+
+
+ For more practice on taking the derivative of a product or quotient,
+
+
+ For more practice on taking the derivative of a composition,
+
+
+
+ Different notations for the derivative
+
+ The difference between
+ \frac{d}{dx} and \frac{dy}{dx} can be confusing. \frac{d}{dx} is a verb and means take the derivative of what comes next;
+ \frac{dy}{dx} is a noun and means the derivative of y with respect to x. Both \frac{dy}{dx} and y' are equivalent notations,
+ but we prefer to use the different forms at different times for reasons that are hard to understand now.
+
+ Treating <m>y</m> as a function of <m>x</m>
+
- In an equation involving x and y where portions of the graph can be defined by explicit functions of x,
- we say that y is an implicit function of x.
- A good example of such a curve is the unit circle.
+ In the problems below, treat y as a function of x, which means that you need
+ to use chain rule to take the derivative with respect to x of expressions involving y.
-
-
-
- We use implicit differentiation to differentiate an implicitly defined function.
- We differentiate both sides of the equation with respect to x,
- treating y as a function of x by applying the chain rule.
- If possible,
- we subsequently solve for \frac{dy}{dx} using algebra.
+ Your answers after taking the derivative will have \frac{dy}{dx} in them, which you will write as y'.
-
-
-
+
+
+
+ Treating <m>y</m> as a function of <m>x</m>
+
- While \frac{dy}{dx} may now involve both the variables x and y,
- \frac{dy}{dx} still gives the slope of the tangent line to the curve.
- It may be used to decide where the tangent line is horizontal (\frac{dy}{dx} = 0) or vertical (\frac{dy}{dx} is undefined),
- or to find the equation of the tangent line at a particular point on the curve.
+ In the problems below, treat y as a function of x, which means that you need
+ to use chain rule to take the derivative with respect to x of expressions involving y.
-
-
-
+
+ Your answers after taking the derivative will have \frac{dy}{dx} in them, which you will write as y'.
+
+
+
+
+
+
-
+ Calculus Practice
+
+
+
+ Implicit differentiation and the tangent line
+
+
+ Implicit differentiation
+
+
+ Dependent versus independent variable
+
+
+ Slope of the curve
+
+
+ Equation of the tangent line
+
+
+ Finding the derivative
+
+
+ Calculating the derivative
+
+
+ Using the tangent line to estimate nearby values
+
+
+ Implicit differentiation and values
+
+
+ Horizontal tangent lines
+
+
+ Horizontal and vertical tangent lines
+
+
+ Implicit differentiation
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/webworkfiles/sec_2-7_prob1.pg b/webworkfiles/sec_2-7_prob1.pg
new file mode 100644
index 0000000..07aad4c
--- /dev/null
+++ b/webworkfiles/sec_2-7_prob1.pg
@@ -0,0 +1,146 @@
+## DESCRIPTION
+##
+## ENDDESCRIPTION
+
+## DBsubject(Algebra)
+## DBchapter(Functions)
+## DBsection(Definition, concept)
+## Date(12/22/2016)
+## Institution(Red Rocks Community College, Colorado Community College System)
+## Author(Brenda Forland)
+## Editor(Chrissy Safranski)
+## Edited date(10/09/2025)
+## MO(1)
+## KEYWORDS('algebra')
+
+
+###########################
+# Initialization
+
+DOCUMENT();
+
+loadMacros(
+"PGstandard.pl",
+"MathObjects.pl",
+"AnswerFormatHelp.pl",
+"parserPopUp.pl",
+"parserImplicitEquation.pl",
+"PGML.pl",
+"PGcourse.pl",
+);
+
+TEXT(beginproblem());
+
+install_problem_grader(~~&std_problem_grader);
+
+$showPartialCorrectAnswers = 0;
+
+
+###########################
+# Setup
+
+Context("Numeric");
+ Context()->variables->add(y => 'Real');
+
+$a=list_random(2,4,6);
+$f1 = Compute("y**$a");
+$popup1 = PopUp(
+ ["?","Explicit function of x","Implicit function of x"],
+ "Implicit function of x",
+);
+$popup1a = PopUp(
+ ["?","Passes vertical line test","Doesn't pass vertical line test"],
+ "Doesn't pass vertical line test",
+);
+
+
+Context()->variables->set(
+x=>{limits=>[-1,1]},
+y=>{limits=>[0,4]}
+);
+$b=non_zero_random(2,8);
+$c=non_zero_random(1,5);
+$f2 = Compute("$b*x**2");
+$popup2 = PopUp(
+ ["?","Explicit function of x","Implicit function of x"],
+ "Explicit function of x",
+);
+
+$popup2a = PopUp(
+ ["?","Passes vertical line test","Doesn't pass vertical line test"],
+ "Passes vertical line test",
+);
+
+
+$d=non_zero_random(2,8);
+$e=non_zero_random(-5,5);
+$f3 = Compute("$d*x");
+$f32 = Compute("y**2");
+$popup3 = PopUp(
+ ["?","Explicit function of x","Implicit function of x"],
+ "Implicit function of x",
+);
+$popup3a = PopUp(
+ ["?","Passes vertical line test","Doesn't pass vertical line test"],
+ "Doesn't pass vertical line test",
+);
+
+
+$f4 = Compute("sqrt(1-x^$b)");
+$popup4 = PopUp(
+ ["?","Explicit function of x","Implicit function of x"],
+ "Explicit function of x",
+);
+$popup4a = PopUp(
+ ["?","Passes vertical line test","Doesn't pass vertical line test"],
+ "Passes vertical line test",
+);
+
+
+
+$h=non_zero_random(-10,-2);
+$f5 = Compute("$h*x*y");
+$popup5 = PopUp(
+ ["?","Explicit function of x","Implicit function of x"],
+ "Explicit function of x",
+);
+$popup5a = PopUp(
+ ["?","Passes vertical line test","Doesn't pass vertical line test"],
+ "Passes vertical line test",
+);
+###########################
+# Main text
+
+BEGIN_PGML
+
+Determine if the following relations represent [`y`] as an explicit or implicit function of [`x`].
+
++ [` x=[$f1] `]
+[___________]{$popup1} [___________]{$popup1a}
+
++ [`[$f2]+y=[$c]`]
+[___________]{$popup2} [___________]{$popup2a}
+
++ [`[$f3]+[$f32]=[$e]`]
+[___________]{$popup3} [___________]{$popup3a}
+
++ [` y=[$f4] `]
+[___________]{$popup4} [___________]{$popup4a}
+
++ [`[$f5]=1`]
+[___________]{$popup5} [___________]{$popup5a}
+
+
+END_PGML
+
+
+############################
+# Solution
+
+#BEGIN_PGML_SOLUTION
+#Solution explanation goes here.
+#END_PGML_SOLUTION
+
+COMMENT('MathObject version. Uses PGML.');
+
+ENDDOCUMENT();
\ No newline at end of file