AP Course

AP Calculus AB/BC

Study the complete College Board sequence for AP Calculus AB and BC, from limits through infinite series.

Choose an official unit and topic to open its lecture, concept check, or focused practice.

Lessons

1.1 Introducing Calculus: Can Change Occur at an Instant?1.2 Defining Limits and Using Limit Notation1.3 Estimating Limit Values from Graphs1.4 Estimating Limit Values from Tables1.5 Determining Limits Using Algebraic Properties of Limits1.6 Determining Limits Using Algebraic Manipulation1.7 Selecting Procedures for Determining Limits1.8 Determining Limits Using the Squeeze Theorem1.9 Connecting Multiple Representations of Limits1.10 Exploring Types of Discontinuities1.11 Defining Continuity at a Point1.12 Confirming Continuity over an Interval1.13 Removing Discontinuities1.14 Connecting Infinite Limits and Vertical Asymptotes1.15 Connecting Limits at Infinity and Horizontal Asymptotes1.16 Working with the Intermediate Value Theorem (IVT)2.1 Defining Average and Instantaneous Rates of Change at a Point2.2 Defining the Derivative of a Function and Using Derivative Notation2.3 Estimating Derivatives of a Function at a Point2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist2.5 Applying the Power Rule2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple2.7 Derivatives of cos x, sin x, e^x, and ln x2.8 The Product Rule2.9 The Quotient Rule2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions3.1 The Chain Rule3.2 Implicit Differentiation3.3 Differentiating Inverse Functions3.4 Differentiating Inverse Trigonometric Functions3.5 Selecting Procedures for Calculating Derivatives3.6 Calculating Higher-Order Derivatives4.1 Interpreting the Meaning of the Derivative in Context4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration4.3 Rates of Change in Applied Contexts Other Than Motion4.4 Introduction to Related Rates4.5 Solving Related Rates Problems4.6 Approximating Values of a Function Using Local Linearity and Linearization4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms5.1 Using the Mean Value Theorem5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points5.3 Determining Intervals on Which a Function Is Increasing or Decreasing5.4 Using the First Derivative Test to Determine Relative (Local) Extrema5.5 Using the Candidates Test to Determine Absolute (Global) Extrema5.6 Determining Concavity of Functions over Their Domains5.7 Using the Second Derivative Test to Determine Extrema5.8 Sketching Graphs of Functions and Their Derivatives5.9 Connecting a Function, Its First Derivative, and Its Second Derivative5.10 Introduction to Optimization Problems5.11 Solving Optimization Problems5.12 Exploring Behaviors of Implicit Relations6.1 Exploring Accumulations of Change6.2 Approximating Areas with Riemann Sums6.3 Riemann Sums, Summation Notation, and Definite Integral Notation6.4 The Fundamental Theorem of Calculus and Accumulation Functions6.5 Interpreting the Behavior of Accumulation Functions Involving Area6.6 Applying Properties of Definite Integrals6.7 The Fundamental Theorem of Calculus and Definite Integrals6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation6.9 Integrating Using Substitution6.10 Integrating Functions Using Long Division and Completing the Square6.11 Integrating Using Integration by Parts6.12 Using Linear Partial Fractions6.13 Evaluating Improper Integrals6.14 Selecting Techniques for Antidifferentiation7.1 Modeling Situations with Differential Equations7.2 Verifying Solutions for Differential Equations7.3 Sketching Slope Fields7.4 Reasoning Using Slope Fields7.5 Approximating Solutions Using Euler’s Method7.6 Finding General Solutions Using Separation of Variables7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables7.8 Exponential Models with Differential Equations7.9 Logistic Models with Differential Equations8.1 Finding the Average Value of a Function on an Interval8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts8.4 Finding the Area Between Curves Expressed as Functions of x8.5 Finding the Area Between Curves Expressed as Functions of y8.6 Finding the Area Between Curves That Intersect at More Than Two Points8.7 Volumes with Cross Sections: Squares and Rectangles8.8 Volumes with Cross Sections: Triangles and Semicircles8.9 Volume with Disc Method: Revolving Around the x- or y-Axis8.10 Volume with Disc Method: Revolving Around Other Axes8.11 Volume with Washer Method: Revolving Around the x- or y-Axis8.12 Volume with Washer Method: Revolving Around Other Axes8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled9.1 Defining and Differentiating Parametric Equations9.2 Second Derivatives of Parametric Equations9.3 Finding Arc Lengths of Curves Given by Parametric Equations9.4 Defining and Differentiating Vector-Valued Functions9.5 Integrating Vector-Valued Functions9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions9.7 Defining Polar Coordinates and Differentiating in Polar Form9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve9.9 Finding the Area of the Region Bounded by Two Polar Curves10.1 Defining Convergent and Divergent Infinite Series10.2 Working with Geometric Series10.3 The nth Term Test for Divergence10.4 Integral Test for Convergence10.5 Harmonic Series and p-Series10.6 Comparison Tests for Convergence10.7 Alternating Series Test for Convergence10.8 Ratio Test for Convergence10.9 Determining Absolute or Conditional Convergence10.10 Alternating Series Error Bound10.11 Finding Taylor Polynomial Approximations of Functions10.12 Lagrange Error Bound10.13 Radius and Interval of Convergence of Power Series10.14 Finding Taylor or Maclaurin Series for a Function10.15 Representing Functions as Power Series

Quizzes

Practice Problems AP formula notes, graph references, and practice sets will be added here.

AP Calculus AB/BC · Unit 9 · Topic 9.2 · BC Only

Second Derivatives of Parametric Equations

Differentiate the parametric tangent slope with respect to x to analyze concavity.

1. Topic Focus

Represent planar motion parametrically and with vectors, then analyze polar derivatives and areas.

This topic: Differentiate the parametric tangent slope with respect to x to analyze concavity.

2. Key Relationship

\(\frac{d^2y}{dx^2}=\frac{d}{dt}(dy/dx)\Big/\frac{dx}{dt}\)

Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.

3. Visual Connection

Parametric derivativesComponent rates determine tangent slope, direction of travel, and concavity along the curve.

4. Worked Example

Differentiate dy/dx with respect to t, then divide by dx/dt again.

Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.

5. Concept Development

BC-only topic. The second derivative measures how the tangent slope changes with respect to \(x\), not with respect to the parameter. Begin with

\(\frac{dy}{dx}=\frac{y'(t)}{x'(t)}.\)

This slope is a function of \(t\). To differentiate it with respect to \(x\), apply the parametric derivative rule one more time:

\(\boxed{\frac{d^2y}{dx^2}=\frac{\dfrac d{dt}\left(\dfrac{dy}{dx}\right)}{dx/dt}}\qquad(x'(t)\ne0).\)

Why divide by \(x'(t)\) again?

Let \(m(t)=dy/dx\). The chain rule gives

\(\frac{dm}{dt}=\frac{dm}{dx}\frac{dx}{dt}=\frac{d^2y}{dx^2}\frac{dx}{dt}.\)

Solving for \(dm/dx\) produces the formula above. Merely differentiating \(dy/dx\) with respect to \(t\) gives the rate at which slope changes per unit parameter, not per unit horizontal change.

Equivalent component formula

Applying the quotient rule to \(y'/x'\) gives a useful compact form:

\(\frac{d^2y}{dx^2}=\frac{x'(t)y''(t)-y'(t)x''(t)}{[x'(t)]^3},\qquad x'(t)\ne0.\)

Either form is valid. The two-step method is often less error-prone: find \(dy/dx\), differentiate that expression with respect to \(t\), then divide by \(x'(t)\).

Concavity

\(\frac{d^2y}{dx^2}>0\Rightarrow\text{concave up},\qquad \frac{d^2y}{dx^2}<0\Rightarrow\text{concave down}.\)

Solve the sign inequality in terms of \(t\), then use the parameter interval to describe the correct pieces of the curve. The formula already accounts for whether \(x\) increases or decreases as \(t\) increases; do not reverse the conclusion when \(x'(t)<0\).

Inflection points

A zero or undefined second derivative is only a candidate. Concavity must actually change as the curve passes through the point. Report the geometric point \((x(t),y(t))\), not only the parameter value. Points where \(x'(t)=0\) require separate geometric or limiting analysis because the standard formula is not valid there.

Reliable procedure

Compute \(x'(t)\) and \(y'(t)\).
Form and simplify \(dy/dx=y'/x'\).
Differentiate that slope with respect to \(t\).
Divide by \(x'(t)\).
State restrictions, especially \(x'(t)\ne0\).
Evaluate the sign or solve sign intervals for concavity.

Incorrect shortcut. In general,

\(\frac{d^2y}{dx^2}\ne\frac{d^2y/dt^2}{d^2x/dt^2}.\)

The right side compares component accelerations and does not differentiate the tangent slope with respect to \(x\).

6. Detailed Worked Example and Error Check

Example 1: Apply the rule twice

For \(x=t^2+1\), \(y=t^3\),

\(\frac{dy}{dx}=\frac{3t^2}{2t}=\frac32t,\qquad \frac{d^2y}{dx^2}=\frac{d(\tfrac32t)/dt}{2t}=\frac{3}{4t},\quad t\ne0.\)

The curve is concave up for \(t>0\) and concave down for \(t<0\).

Example 2: A rational first derivative

For \(x=t^2-3\), \(y=2t-1\),

\(\frac{dy}{dx}=\frac1t,\qquad \frac{d^2y}{dx^2}=\frac{-1/t^2}{2t}=-\frac1{2t^3},\quad t\ne0.\)

At \(t=1\) the curve is concave down; at \(t=-1\) it is concave up.

Example 3: Concavity of a circle

Let \(x=4\cos t\), \(y=4\sin t\). Then \(dy/dx=-\cot t\), so

\(\frac{d^2y}{dx^2}=\frac{\csc^2t}{-4\sin t}=-\frac1{4\sin^3t}.\)

At \(t=\pi/4\), the value is \(-1/\sqrt2\), so the upper-right part of the circle is concave down.

Example 4: Concavity intervals and an inflection point

For \(x=t\), \(y=t^3-3t\),

\(\frac{dy}{dx}=3t^2-3,\qquad \frac{d^2y}{dx^2}=6t.\)

The curve is concave down for \(t<0\) and concave up for \(t>0\). Concavity changes at \(t=0\), so \((0,0)\) is an inflection point.

Example 5: A removable parameter restriction

For \(x=t^2\), \(y=t^4\), the curve is \(y=x^2\) with \(x\ge0\). For \(t\ne0\),

\(\frac{dy}{dx}=\frac{4t^3}{2t}=2t^2,\qquad \frac{d^2y}{dx^2}=\frac{4t}{2t}=2.\)

The parametric formula is initially undefined at \(t=0\), where \(x'=0\), but the rectangular equation shows that the second derivative extends continuously to \(2\) at the origin.

Example 6: Reversing the parameter direction

The equations \(x=-t\), \(y=t^2\) trace the parabola \(y=x^2\) with \(x\) decreasing as \(t\) increases. Nevertheless,

\(\frac{dy}{dx}=\frac{2t}{-1}=-2t,\qquad \frac{d^2y}{dx^2}=\frac{-2}{-1}=2.\)

The curve remains concave up. No extra sign reversal is needed.

Example 7: Use component derivative data

At \(t=t_0\), suppose \(x'=2\), \(y'=3\), \(x''=-1\), and \(y''=4\). The compact formula gives

\(\frac{d^2y}{dx^2}=\frac{(2)(4)-(3)(-1)}{2^3}=\frac{11}{8}>0.\)

The curve is concave up at the corresponding point.

Example 8: Zero does not guarantee inflection

For \(x=t\), \(y=(t-1)^4\),

\(\frac{d^2y}{dx^2}=12(t-1)^2.\)

The second derivative is zero at \(t=1\), but it is nonnegative on both sides. Concavity does not change, so \((1,0)\) is not an inflection point.

Common errors

Stopping after \(d(dy/dx)/dt\) and forgetting to divide by \(dx/dt\).
Using \(y''(t)/x''(t)\).
Differentiating \(y'(t)/x'(t)\) incorrectly with the quotient rule.
Canceling a factor without recording excluded parameter values.
Using the formula at \(x'(t)=0\) without separate analysis.
Reversing concavity because the curve is traced right to left.
Calling every zero of the second derivative an inflection point.
Reporting only \(t\) when a point is requested.

7. AP Reasoning Routine

Keep the parameter visible until the requested quantity is formed, track orientation and speed, and choose polar bounds from the traced region.

Identify the representation and requested quantity.
State the rule or theorem and verify its conditions.
Keep exact values until the final requested approximation.
Interpret sign, units, interval, and context.

AP Checkpoint

Find the second derivative and answer the requested concavity question.
(a) For \(x=t^2+1\), \(y=t^3\), find \(d^2y/dx^2\) and classify concavity at \(t=2\).
(b) For \(x=t^2-3\), \(y=2t-1\), find \(d^2y/dx^2\) and classify concavity at \(t=-1\).
(c) For \(x=\cos t\), \(y=\sin t\), find \(d^2y/dx^2\) and evaluate it at \(t=\pi/4\).
(d) For \(x=t\), \(y=t^3-3t\), determine the concavity intervals and any inflection point.
(e) For \(x=e^t\), \(y=e^{-t}\), find \(d^2y/dx^2\) and determine its sign.
(f) For \(x=t^2\), \(y=t^4\), find \(d^2y/dx^2\) where the parametric formula applies and explain what happens at \(t=0\).
(g) At \(t=t_0\), suppose \(x'=2\), \(y'=-1\), \(x''=3\), and \(y''=4\). Find \(d^2y/dx^2\).
(h) For \(x=t\), \(y=t^4\), decide whether \(t=0\) is an inflection point.
(i) Explain why \(y''(t)/x''(t)\) is not generally \(d^2y/dx^2\).
(j) Explain why the standard second-derivative formula cannot be applied directly at a vertical tangent.

Check the solution

(a) \(dy/dx=3t/2\), so \(d^2y/dx^2=3/(4t)\). At \(t=2\), it is \(3/8>0\), so the curve is concave up.
(b) \(d^2y/dx^2=-1/(2t^3)\). At \(t=-1\), it is \(1/2>0\), so the curve is concave up.
(c) \(dy/dx=-\cot t\), hence \(d^2y/dx^2=-\csc^2t/\sin t=-1/\sin^3t\). At \(t=\pi/4\), the value is \(-2\sqrt2\).
(d) \(d^2y/dx^2=6t\). The curve is concave down for \(t<0\), concave up for \(t>0\), and has an inflection point at \((0,0)\).
(e) \(dy/dx=-e^{-2t}\), so \(d^2y/dx^2=2e^{-3t}>0\) for every \(t\).
(f) For \(t\ne0\), \(dy/dx=2t^2\) and \(d^2y/dx^2=2\). At \(t=0\), \(x'=0\), but eliminating the parameter gives \(y=x^2\), so the value extends to \(2\).
(g) Using \([x'y''-y'x'']/(x')^3\), the value is \([2(4)-(-1)(3)]/8=11/8\).
(h) \(d^2y/dx^2=12t^2\). It is zero at \(t=0\) but positive on both sides, so there is no change in concavity and no inflection point.
(i) The second derivative differentiates the slope \(dy/dx\) with respect to \(x\). Component second derivatives measure change with respect to \(t\) and their quotient does not perform that operation.
(j) A vertical tangent has \(x'(t)=0\), making the formula divide by zero. Concavity there requires a different representation or one-sided geometric analysis.