For a parametric curve, dy/dx is (dy/dt) ÷ (dx/dt). The second derivative is the derivative of dy/dx with respect to t, divided again by dx/dt. Speed and arc length both use √((dx/dt)² + (dy/dt)²), Pythagoras applied to an infinitesimal step.
Step 1: Picture It
Start with something you already know.
A robot on a tiled floor
- A robot rolls across a big floor of square tiles.
- Each second it goes 5 tiles east and 12 tiles north.
- Its track climbs 12 tiles for every 5 across. That steepness is the slope, 12 ÷ 5.
- Its real move each second is the long side of a right triangle.
- That side is √(25 + 144) = 13 tiles, so its speed is 13 tiles a second.
The same two rates give two different answers. Divide them for the slope. Use Pythagoras on them for the speed.
On a curve, the rates change from moment to moment. So the slope and the speed are read one moment at a time, from that moment’s rates.
Where the picture falls short: this robot never changes its rates, so its distance is just speed times time. On a curve the speed changes, so the distance adds the speeds up with an integral. That integral is the arc length.
Step 2: Let's Learn
Read it, or press Listen and follow the words.
Try it first
A point moves with x = 2t and y = t². Find where it is at t = 0, t = 1 and t = 2. Then add the two straight-line distances: from the first point to the second, and from the second to the third. Keep the full decimals and round only the total, to one decimal place.
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
Why it works
The chain rule gives dy/dt = (dy/dx)(dx/dt). Rearranging isolates dy/dx without ever eliminating t. Check: x = t², y = t³ is y = x^(3/2), with y′ = (3/2)√x = (3/2)t. It agrees.
The second derivative
d²y/dx² = (d/dt of dy/dx) ÷ (dx/dt). For the curve above: dy/dx = 3t/2, its t-derivative is 3/2, divided by 2t gives 3/(4t). Dividing only once, getting 3/2, is the standard error.
Reading the second derivative
3/(4t) is positive for t > 0: concave up there. Negative for t < 0: concave down. The Cartesian check: y = x^(3/2) has y″ = (3/4)x^(−1/2) = 3/(4t). Same.
Vertical and horizontal tangents
Vertical where dx/dt = 0 and dy/dt ≠ 0. Horizontal the reverse. At t = 0 both are zero here: a cusp, and the formula says nothing. Check dx/dt before dividing.
Arc length
L = ∫ √((dx/dt)² + (dy/dt)²) dt between the t limits. For x = t², y = t³ from t = 0 to 1: ∫₀¹ √(4t² + 9t⁴) dt = ∫₀¹ t√(4 + 9t²) dt. Substituting u = 4 + 9t²: (1/18)(⅔)(13^(3/2) − 8) ≈ 1.44.
Speed
That same square root is the speed of the particle at time t: √(4t² + 9t⁴). At t = 1: √13 ≈ 3.61. Integrating speed over time gives distance traveled, which is the arc length.
Step 3: Try It Yourself
Tap and try it out.
x = t², y = t³. Find d²y/dx², and the arc length from t = 0 to t = 1
- 1dy/dx = 3t²/(2t) = 3t/2the first derivative
Step 0 of 5
- Point(2, 4)
- Slope of the tangent4
Step 4: In Real Life
Miles from a GPS track
From x(t) and y(t), the speed is √(x′² + y′²) and the distance is its integral: the arc length. That is how a phone turns a GPS track into miles run.
Step 5: Watch an Example
One step at a time.
Watch Noor Find a Slope and a Bend
Noor has x = t² + 1 and y = t³ − 3t. She wants the slope and the second derivative at t = 1, and what they say about the curve there.
- Step 1
The two rates come first: dx/dt = 2t and dy/dt = 3t² − 3. At t = 1, dx/dt = 2. That is not zero, so dividing by it is safe there.
Watch Sam Measure a Path, Then Finish It Yourself
A point moves with x = 3t² and y = t³ − 3t. Sam finds its speed formula. The speed at one moment, the length from t = 0 to t = 1, and a check on that length are yours.
- Step 1
The rates are dx/dt = 6t and dy/dt = 3t² − 3. The speed is the square root of the sum of their squares.
Watch Ana Find the Horizontal and Vertical Tangents
Ana has x = t² − 2t and y = t³ − 12t. She wants every point where the tangent is vertical or horizontal.
- Step 1
The rates come first: dx/dt = 2t − 2 and dy/dt = 3t² − 12.
Step 6: Your Turn
Practice makes it stick.
The Survey Drone
Problem 1 of 15
A drone flies over a field. After t seconds it is at x = 6t meters east and y = t² + 2 meters north of a fence post. At what time t, in seconds, is the drone’s speed 10 meters per second?
seconds
The Test Track
Problem 2 of 15
A toy car drives a curved test track. After t seconds it is at x = t² meters and y = t³/3 − t meters. How far does it travel along the track from t = 0 to t = 3 seconds, in meters?
meters
Read It From the Rates
Find the two rates first every time. Then decide what to do with them.
Problem 3 of 15
x = t³ + t and y = t² + 4t. What is dy/dx at t = 1? Give it as a decimal.
Problem 4 of 15
x = 4t and y = t². What is the speed at t = 1.5?
Problem 5 of 15
x = 2t and y = t³. What is d²y/dx² at t = 2?
Problem 6 of 15
x = t³ + 1 and y = t² − 6t. At what value of t is the tangent horizontal?
Problem 7 of 15
x = 3t² and y = 2t³. What is the arc length from t = 0 to t = 1? Round to two decimal places.
Whose Step?
One line on each page is a slip. Find it, then put it right.
Problem 8 of 15
Ravi worked out d²y/dx² at t = 2 for x = 4t and y = t³ − t. One line is a slip. Which?
Ravi’s working for d²y/dx² at t = 2, where x = 4t and y = t³ − t
- 1dx/dt = 4, dy/dt = 3t² − 1
- 2dy/dx = (3t² − 1)/4
- 3d/dt of (3t² − 1)/4 = 6t/4 = 1.5t
- 4d²y/dx² = 1.5t
- Answerat t = 2: d²y/dx² = 3
Problem 9 of 15
Done correctly, what is d²y/dx² at t = 2 for x = 4t and y = t³ − t? Give it as a decimal.
Problem 10 of 15
Leo looked for vertical tangents on x = t² − 4t, y = t³ − 3t. One line is a slip. Which?
Leo’s search for vertical tangents on x = t² − 4t, y = t³ − 3t
- 1dx/dt = 2t − 4, dy/dt = 3t² − 3
- 2a vertical tangent needs dy/dt = 0
- 33t² − 3 = 0 at t = 1 and t = −1
- Answervertical tangents at t = 1 and t = −1
Problem 11 of 15
For x = t² − 4t and y = t³ − 3t, at what value of t is the tangent really vertical?
Pick the Move
Choose the move that answers the question. Neither of these needs working all the way through.
Problem 12 of 15
The slope dy/dx is wanted at t = 1 on x = t³ − 3t, y = t². What should come first?
Problem 13 of 15
The length of the path x = 3t, y = t² from t = 0 to t = 2 is wanted. Which move finds it?
Same Slope, Same Speed?
Work out the two rates at t = 1 for each one. Select every entry that fits.
Problem 14 of 15
Select every motion whose speed at t = 1 is 5.
Problem 15 of 15
Select every curve whose slope dy/dx at t = 1 is 2.
Step 7: Quick Check
Show what you know.
Question 1 of 2
dy/dt = 18 and dx/dt = 6. What is dy/dx?
Question 2 of 2
What is the common error when finding the second derivative?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
x = t² + kt and y = t³ − 12t. For which number k are both rates zero at the same positive t, so the curve has a cusp there?
Stretch 2 of 2
x = 6t² and y = t³ − ct, with c a positive number. For exactly one c the sum of the squared rates is a perfect square, so the speed has no root left in it. Find that c, then type the arc length from t = 0 to t = 1.
What You Learned
- dy/dx is (dy/dt) ÷ (dx/dt).
- The second derivative divides by dx/dt a second time.
- Speed and arc length both use √((dx/dt)² + (dy/dt)²).
- Check dx/dt before dividing: where it is zero the tangent is vertical.
For the grown-up
The misconception this lesson targets is that the second derivative of a parametric curve is just the derivative of dy/dx with respect to t. It looks right, because the first derivative came from a single division, and the student stops there. But dy/dx is written in t, and d²y/dx² measures change per unit of x, so a second division by dx/dt is owed. A close cousin is mixing up the two things built from the same pair of rates: their ratio is the slope, which says which way the path heads, and Pythagoras on them is the speed, which says how fast. The habits that settle both are cheap: find the two rates first, check dx/dt before dividing, and when a second derivative is asked for, count the divisions.