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Amy Ferguson Moncure

Calculus BC · Unit 9

Second Derivatives and Parametric Arc Length

Differentiate the slope with respect to t and divide by dx/dt. Arc length integrates the speed.

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Key Ideas Easy Medium Hard

\(L=\int_a^b \sqrt{(dx/dt)^2+(dy/dt)^2}\,dt\)

What to Remember

Second derivative

\(\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{dx/dt}\)

Differentiate the first slope with respect to \(t\), then divide by \(dx/dt\) again.

Concavity follows the sign

The curve is concave up where the second derivative is positive, at a parameter value where the formula applies.

Parametric arc length

\(L=\int_a^b \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt\)

Speed is the integrand

The expression under the square root, after the root is taken, is the speed. Distance traveled is the integral of speed.

Level 1

Easy

Speed is constant, so the length is speed times time.

Example
\(x=3t\) and \(y=4t\) for \(0\le t\le 2\). Find the length of the curve.

\(\dfrac{dx}{dt}=3\) and \(\dfrac{dy}{dt}=4\).

\(\sqrt{3^2+4^2}\)

\(5\)

\(\int_0^2 5\,dt\)

\(10\)

Answer: \(10\)

1
\(x=2\cos t\) and \(y=2\sin t\) for \(0\le t\le \pi/2\). Find the arc length.
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Speed: \(\sqrt{(-2\sin t)^2+(2\cos t)^2}=2\).

\(\int_0^{\pi/2} 2\,dt\)

\(\pi\)

This is a quarter circle of radius 2.

Answer: \(\pi\)

2
Find the speed of \(x=t^2\), \(y=t^3-3t\) at \(t=2\).
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\(dx/dt=4\) and \(dy/dt=9\).

\(\sqrt{16+81}\)

\(\sqrt{97}\)

Answer: \(\sqrt{97}\)

3
\(x=5\cos t\) and \(y=5\sin t\). Find the speed at every \(t\).
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\(\sqrt{25\sin^2 t+25\cos^2 t}\)

\(5\)

Answer: \(5\)

4
A particle follows \(x=t\), \(y=t\) from \(t=1\) to \(t=4\). How far does it travel?
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Speed \(\sqrt{1+1}=\sqrt{2}\).

\(3\sqrt{2}\)

Answer: \(3\sqrt{2}\)

5
Which integral is the length for \(a\le t\le b\)?
A. \(\int_a^b \sqrt{(dx/dt)^2+(dy/dt)^2}\,dt\)
B. \(\int_a^b \sqrt{1+(dy/dx)^2}\,dt\) with no change of variable
C. \(x(b)-x(a)\)
D. \(\int_a^b (dx/dt+dy/dt)\,dt\)
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The parameter is \(t\), so both component rates belong under the radical.

The graph formula in \(x\) is a different setup.

Answer: A. \(\int_a^b\sqrt{(x')^2+(y')^2}\,dt\)

Level 2

Medium

Differentiate the first slope with respect to t, then divide by dx/dt.

Example
For \(x=t^2\) and \(y=t^3-3t\), \(dy/dx=(3t^2-3)/(2t)\). Find \(d^2y/dx^2\) at \(t=2\).

Differentiate \((3t^2-3)/(2t)\) with respect to \(t\).

\(\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)=\dfrac{3(t^2+1)}{2t^2}\)

At \(t=2\): \(\dfrac{3\cdot 5}{8}=\dfrac{15}{8}\).

Divide by \(dx/dt=4\).

\(\dfrac{15/8}{4}\)

\(\dfrac{15}{32}\)

Answer: \(\dfrac{15}{32}\)

1
At \(t=2\) the second derivative just found is positive. What does that say about concavity?
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A positive second derivative means concave up.

This is at the point \((4,2)\).

Answer: concave up at \((4,2)\)

2
\(x=t\) and \(y=t^3\). Find \(d^2y/dx^2\) at \(t=1\).
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\(dy/dx=3t^2\)

Derivative with respect to \(t\): \(6t\).

Divide by \(dx/dt=1\).

At \(t=1\): \(6\).

Answer: \(6\)

3
Find the length of \(x=\cos t\), \(y=\sin t\) from \(t=0\) to \(t=\pi\).
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Speed is 1.

\(\int_0^{\pi} 1\,dt\)

\(\pi\)

Answer: \(\pi\)

4
\(dx/dt=t\) and \(dy/dt=t\) for \(1\le t\le 3\). Find the distance traveled.
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Speed \(\sqrt{t^2+t^2}=t\sqrt{2}\) since \(t>0\).

\(\int_1^3 t\sqrt{2}\,dt\)

\(\sqrt{2}\left[\dfrac{t^2}{2}\right]_1^3\)

\(\sqrt{2}\cdot 4\)

\(4\sqrt{2}\)

Answer: \(4\sqrt{2}\)

5
Why is \(d^2y/dx^2\) not equal to \((d^2y/dt^2)/(d^2x/dt^2)\)?
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The second derivative in \(x\) is the derivative of \(dy/dx\), not the ratio of the two second derivatives in \(t\).

You differentiate the slope with respect to \(t\), then divide by \(dx/dt\).

Answer: differentiate the slope first

Level 3

Hard

A second derivative that changes sign, or a length that simplifies.

Example
For \(x=t\) and \(y=t^3\), determine the concavity on \(t<0\) and on \(t>0\).

\(dy/dx=3t^2\)

\(d^2y/dx^2=6t\)

For \(t<0\), the second derivative is negative: concave down.

For \(t>0\), it is positive: concave up.

Answer: down for \(t<0\), up for \(t>0\)

1
Find the length of \(x=t^2\), \(y=\dfrac{2}{3}t^3\) from \(t=0\) to \(t=\sqrt{3}\). \(dx/dt=2t\) and \(dy/dt=2t^2\).
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\(\sqrt{4t^2+4t^4}=2|t|\sqrt{1+t^2}\)

On \([0,\sqrt{3}]\), \(|t|=t\).

This length is \(\int_0^{\sqrt{3}} 2t\sqrt{1+t^2}\,dt\).

Let \(u=1+t^2\). Then \(du=2t\,dt\), from 1 to 4.

\(\int_1^4 \sqrt{u}\,du\)

\(\left[\dfrac{2}{3}u^{3/2}\right]_1^4\)

\(\dfrac{2}{3}(8-1)\)

\(\dfrac{14}{3}\)

Answer: \(\dfrac{14}{3}\)

2
\(x=2t\) and \(y=t^2\). Find \(d^2y/dx^2\).
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\(dy/dx=\dfrac{2t}{2}=t\)

But \(t=x/2\), so \(dy/dx=x/2\).

Derivative with respect to \(t\) of the slope \(t\) is 1.

Divide by \(dx/dt=2\).

\(\dfrac{1}{2}\)

Answer: \(\dfrac{1}{2}\)

3
A particle moves with \(dx/dt=\cos t\) and \(dy/dt=\sin t\) from \(t=0\) to \(t=\pi/2\). Find the distance traveled.
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Speed \(\sqrt{\cos^2 t+\sin^2 t}=1\).

\(\pi/2\)

Answer: \(\dfrac{\pi}{2}\)

4
At \(t=1\), \(dy/dx=0\) and \(d^2y/dx^2=-4\). Describe the tangent and the concavity.
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Slope 0 means a horizontal tangent.

A negative second derivative means concave down.

The curve has a local maximum shape at that point, if the second derivative test’s other conditions hold in \(x\).

Answer: horizontal tangent, concave down

5
Find the length of \(x=2\cos t\), \(y=2\sin t\) for one full turn, \(0\le t\le 2\pi\).
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Speed is 2.

\(\int_0^{2\pi} 2\,dt\)

\(4\pi\)

The curve is a circle of radius 2, whose circumference is \(2\pi\cdot 2\).

Answer: \(4\pi\)

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