Second Derivatives of Parametric Equations: Study Guide (Calculus BC 9.2)

BC-only material. Parametric equations are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

You need these earlier ideas. If any is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
dy/dx = (dy/dt) ÷ (dx/dt)	9.1	The first step, and the same idea used again
Quotient rule and chain rule	2.9, 3.1	dy/dx is often a quotient of functions of t
Second derivatives and concavity	3.6, 5.6	Reading the sign of d²y/dx²
Tangent-line approximation	4.6	Deciding if an estimate is too big or too small

Notation on this page: x′(t) = dx/dt, y′(t) = dy/dt, and x″(t), y″(t) are the second derivatives with respect to t. Angles are in radians. You should already be confident with Topic 9.1.

The tempting shortcut that fails

For y = f(x), the second derivative d²y/dx² measures how the slope changes as x changes. Its sign gives the concavity.

For a parametric curve, it is tempting to differentiate x and y twice with respect to t and divide: d²y/dt² ÷ d²x/dt². This is wrong. Here is a quick test.

Take x = 2t, y = t². Eliminating t gives y = x²/4, so the true second derivative is d²y/dx² = ½ everywhere. But d²y/dt² = 2 and d²x/dt² = 0, so the shortcut gives 2 ÷ 0, which is undefined. The shortcut fails even on a simple parabola.

The reason: d²y/dx² means “the derivative with respect to x of dy/dx”. Nothing in the shortcut takes a derivative with respect to x.

Deriving the correct method

Call the slope m = dy/dx. In a parametric problem, m is a function of t. We want dm/dx, the rate at which the slope changes as x changes.

This is exactly the problem from Topic 9.1, with m in place of y. The chain rule gives dm/dt = (dm/dx)(dx/dt), so, provided dx/dt ≠ 0,

d²y/dx² = dm/dx = [d/dt (dy/dx)] ÷ (dx/dt)

Check with x = 2t, y = t²: dy/dx = 2t/2 = t. Then d/dt (t) = 1, and 1 ÷ (dx/dt) = 1 ÷ 2 = ½. ✓ This matches y = x²/4.

The method in four steps.

1. Find dx/dt and dy/dt.
2. Find dy/dx = (dy/dt) ÷ (dx/dt), and simplify it as a function of t.
3. Differentiate dy/dx with respect to t.
4. Divide the result by dx/dt.

Simplifying in step 2 saves a lot of quotient-rule work in step 3.

Background: a single formula. If you apply the quotient rule to (y′/x′) and divide by x′, you get d²y/dx² = (x′y″ − y′x″) ÷ (x′)³. It gives the same answer and can be used as a check. You do not need to memorise it.

What the sign tells you

Once you have d²y/dx², you read it exactly as for y = f(x):

Sign of d²y/dx²	Shape near the point	Tangent line nearby	Tangent-line estimate
Positive	Concave up	Lies below the curve	Underestimate
Negative	Concave down	Lies above the curve	Overestimate

Watch the sign of dx/dt. In step 4 you divide by dx/dt. If dx/dt < 0 (the point is moving left), dividing flips the sign. So d/dt (dy/dx) > 0 does not always mean concave up. Worked example 2 shows this happening.

Worked example 1: a second derivative at a point

Question. A curve is given by x = 4 sin t, y = 2 cos 2t. Without a calculator, find d²y/dx² at t = π/6 and state the concavity there.

1. First derivatives. dx/dt = 4 cos t. dy/dt = −4 sin 2t (chain rule: d/dt [cos 2t] = −2 sin 2t).
2. dy/dx, simplified. Use sin 2t = 2 sin t cos t: dy/dx = (−4 sin 2t)/(4 cos t) = (−8 sin t cos t)/(4 cos t) = −2 sin t, for cos t ≠ 0.
3. Differentiate with respect to t. d/dt (−2 sin t) = −2 cos t.
4. Divide by dx/dt. d²y/dx² = (−2 cos t)/(4 cos t) = −½, for cos t ≠ 0.
5. At t = π/6. The point is (4 · ½, 2 cos(π/3)) = (2, 1). The slope is −2 sin(π/6) = −1, and d²y/dx² = −½ < 0, so the curve is concave down there.

Check by eliminating t. cos 2t = 1 − 2 sin²t, so y = 2 − 4 sin²t = 2 − x²/4. Then dy/dx = −x/2 = −2 sin t ✓ and d²y/dx² = −½ ✓. The curve is part of a downward parabola.

The wrong shortcut here. d²y/dt² = −8 cos 2t = −4 and d²x/dt² = −4 sin t = −2 at t = π/6, giving (−4) ÷ (−2) = 2. That has the wrong sign and the wrong size, and would wrongly say “concave up”.

Worked example 2: concavity on intervals, and a tangent-line estimate

Question. A curve is given by x = t² + 2t and y = t³ − 3t (Figure 1). (a) Find dy/dx and d²y/dx² in terms of t. (b) For which values of t is the curve concave up? Concave down? (c) Use the tangent line at t = 0 to estimate y when x = 0.2 on the same part of the curve. Is the estimate too big or too small?

(a) First and second derivatives.

1. dx/dt = 2t + 2 = 2(t + 1). dy/dt = 3t² − 3 = 3(t − 1)(t + 1).
2. dy/dx = 3(t − 1)(t + 1) ÷ 2(t + 1) = 3(t − 1)/2, for t ≠ −1.
3. d/dt (dy/dx) = 3/2. This is positive for every t.
4. d²y/dx² = (3/2) ÷ 2(t + 1) = 3 / (4(t + 1)), for t ≠ −1.

(b) Concavity.

• For t > −1: t + 1 > 0, so d²y/dx² > 0 and the curve is concave up.
• For t < −1: t + 1 < 0, so d²y/dx² < 0 and the curve is concave down.
• At t = −1, dx/dt = 0 and dy/dt = 0, so dy/dx and d²y/dx² are not defined there. Figure 1 shows a sharp point (a cusp) at (−1, 2).

Notice that d/dt (dy/dx) = 3/2 is positive for all t, yet the curve is concave down for t < −1. The difference comes from dividing by dx/dt, which is negative when t < −1 (the point is moving left).

(c) Tangent-line estimate.

1. At t = 0: the point is (0, 0), the slope is 3(0 − 1)/2 = −3/2, and d²y/dx² = 3/4.
2. Tangent line: y = −1.5x. At x = 0.2 it gives y ≈ −0.3.
3. d²y/dx² = 3/4 > 0, so the curve is concave up near t = 0. The tangent line lies below the curve, so the estimate −0.3 is an underestimate.

Checks.

• Actual value: x = 0.2 gives t² + 2t − 0.2 = 0, so t = −1 + √1.2 ≈ 0.0954 on this part of the curve. Then y ≈ −0.2855, which is greater than −0.3. ✓ Underestimate.
• Why “same part” matters: x = 0.2 also occurs at t = −1 − √1.2 ≈ −2.095 on the concave-down part, where y ≈ −2.91. The tangent line at t = 0 says nothing about that point.
• Optional formula: (x′y″ − y′x″)/(x′)³ = [2(t + 1) · 6t − 3(t² − 1) · 2] / [8(t + 1)³] = 6(t + 1)² / [8(t + 1)³] = 3/(4(t + 1)). ✓

Worked example 3: when you are given only dx/dt and dy/dt

Many questions do not give x(t) and y(t) at all. They give only the two rates. That is enough, because the method never uses x or y themselves.

Question. A curve has dx/dt = t² − 4 and dy/dt = 2t. Without a calculator: (a) find d²y/dx² in terms of t; (b) find d²y/dx² at t = 1 and state the concavity there; (c) find the intervals of t on which the curve is concave up and concave down.

(a) Follow the four steps.

1. dx/dt = t² − 4 and dy/dt = 2t are given.
2. dy/dx = 2t/(t² − 4), for t ≠ ±2. Nothing cancels, so you need the quotient rule next.
3. d/dt (dy/dx) = [2(t² − 4) − 2t · 2t] ÷ (t² − 4)² = (−2t² − 8)/(t² − 4)² = −2(t² + 4)/(t² − 4)².
4. Divide by dx/dt = t² − 4: d²y/dx² = −2(t² + 4)/(t² − 4)³, for t ≠ ±2.

(b) At t = 1. dx/dt = −3 and d/dt (dy/dx) = −2(5)/9 = −10/9. So d²y/dx² = (−10/9) ÷ (−3) = 10/27 > 0, and the curve is concave up there. Notice the two negatives: dividing by a negative dx/dt flipped the sign.

(c) Sign analysis. The numerator −2(t² + 4) is always negative. So the sign of d²y/dx² is the opposite of the sign of (t² − 4)³, which has the same sign as t² − 4.

• For −2 < t < 2: t² − 4 < 0, so d²y/dx² > 0 and the curve is concave up.
• For t < −2 or t > 2: t² − 4 > 0, so d²y/dx² < 0 and the curve is concave down.

Checks. At t = 0, the formula gives −8/(−64) = 1/8 > 0 ✓, and at t = 3 it gives −26/125 < 0 ✓. At t = ±2, dx/dt = 0 but dy/dt = ±4 ≠ 0, so the tangent is vertical there (Topic 9.1) and d²y/dx² is not defined. The concavity changes at these points even though d²y/dx² is never zero. So when you look for changes in concavity on a parametric curve, check the values of t where dx/dt = 0 as well as where d²y/dx² = 0.

Common misconceptions

• “d²y/dx² = (d²y/dt²) ÷ (d²x/dt²).” This shortcut is wrong. It fails for x = 2t, y = t², and gives the wrong sign in Worked example 1.
• Stopping after step 3. d/dt (dy/dx) is not the answer. You must divide by dx/dt again.
• Differentiating dy/dx with respect to x when it is written in t. You cannot treat t as x. Differentiate with respect to t, then convert by dividing by dx/dt.
• Reading the concavity from d/dt (dy/dx). When dx/dt < 0 its sign is the opposite of the concavity, as in Worked example 2.
• Not simplifying dy/dx first. A messy quotient makes step 3 long and error-prone. Cancel common factors and use identities such as sin 2t = 2 sin t cos t.
• Ignoring where dx/dt = 0. Neither dy/dx nor d²y/dx² is defined there by this method, yet the concavity can change there, as in Worked example 3.
• Mixing up over and under. Concave down means the tangent line is above the curve, so the estimate is too big.

Where this leads

Topic 9.3 uses dx/dt and dy/dt to find the arc length of a parametric curve. Topics 9.4 to 9.6 treat the parameter as time: there, the second derivatives x″(t) and y″(t) are the components of acceleration, which is a different quantity from d²y/dx². Keep the two separate. Next: Finding Arc Lengths of Curves Given by Parametric Equations (9.3).

Try the practice questions now, then use the revision notes and the checklist to consolidate.