Calculus 2 Quiz: Arc Length Cartesian

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Arc Length CartesianQuestion 1 of 20

The arc length of a continuously differentiable function $f(x)$ from $x=a$ to $x=b$ is given by $L = \int_a^b \sqrt{1+[f'(x)]^2} \, dx$ . If $L = b-a$ , what must be true about the function $f(x)$ on the interval $[a,b]$ ?

f(x)

is a linear function with slope 1.

f(x)

is a constant function.

f(x)

is the identity function,

f(x)=x

f(x)

must be the zero function,

f(x)=0

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Calculus 2 Quiz

Calculus 2 Quiz: Arc Length Cartesian

Practice Arc Length Cartesian in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Arc Length Cartesian, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

$f(x)$ is a linear function with slope 1.
$f(x)$ is a constant function. (correct answer)
$f(x)$ is the identity function, $f(x)=x$ .
$f(x)$ must be the zero function, $f(x)=0$ .

Explanation: If

L = b-a

, then

\int_a^b \sqrt{1+[f'(x)]^2} \, dx = b-a = \int_a^b 1 \, dx

. This implies that the integrand

\sqrt{1+[f'(x)]^2}

must be equal to 1 for all

x

[a,b]

. For this to be true,

1+[f'(x)]^2 = 1

, which means

[f'(x)]^2 = 0

. Therefore,

f'(x) = 0

for all

x

[a,b]

. A function whose derivative is zero on an interval must be a constant function on that interval.

Question 2

The length of the curve $y=x^4$ from $x=0$ to $x=1$ is given by $L_1$ . The length of the curve $y=x^{1/4}$ from $x=0$ to $x=1$ is given by $L_2$ . What is the relationship between $L_1$ and $L_2$ ?

$L_1 > L_2$
$L_1 < L_2$
$L_1 = L_2$ (correct answer)
The relationship cannot be determined without calculating the integrals.

Explanation: The functions

y=x^4

and

y=x^{1/4}

are inverses of each other. The graph of

y=x^{1/4}

for

x \in [0,1]

is a reflection of the graph of

y=x^4

for

x \in [0,1]

across the line

y=x

. Since reflection is a rigid transformation, it does not change the arc length. Therefore, the arc lengths of the two curves over their respective intervals from

(0,0)

(1,1)

must be equal.

L_1 = \int_0^1 \sqrt{1+(4x^3)^2}dx

and

L_2 = \int_0^1 \sqrt{1+(\frac{1}{4}x^{-3/4})^2}dx

. By symmetry, these integrals are equal.

Question 3

Consider the curves $C_1: y=x^2$ and $C_2: y=2x^2$ on the interval $[0,1]$ . Let $L_1$ and $L_2$ be their respective arc lengths. Which statement is correct?

$L_2 = 2L_1$
$L_2 > L_1$ (correct answer)
$L_2 < L_1$
$L_2 = \sqrt{2} L_1$

Explanation: For

C_1

y'=2x

, so

L_1 = \int_0^1 \sqrt{1+4x^2}dx

. For

C_2

y'=4x

, so

L_2 = \int_0^1 \sqrt{1+16x^2}dx

. For any

x

(0,1]

, we have

16x^2 > 4x^2

, which implies

1+16x^2 > 1+4x^2

, and therefore

\sqrt{1+16x^2} > \sqrt{1+4x^2}

. Since the integrand for

L_2

is strictly greater than the integrand for

L_1

over the interval of integration (except at

x=0

), the value of the integral

L_2

must be strictly greater than

L_1

Question 4

A flexible cable is hanging between two poles. Its shape can be modeled by the function $y = 20\cosh(x/20)$ for $-10 \le x \le 10$ . What is the length of the cable between these two x-values?

$40\sinh(1/2)$ (correct answer)
$20\sinh(1/2)$
$40\cosh(1/2)$
$20$

Explanation: The length of the cable is the arc length. First, find the derivative:

y' = 20 \sinh(x/20) \cdot (1/20) = \sinh(x/20)

. Next, compute the integrand

\sqrt{1+(y')^2} = \sqrt{1+\sinh^2(x/20)}

. Using the identity

1+\sinh^2(u) = \cosh^2(u)

, this simplifies to

\sqrt{\cosh^2(x/20)} = \cosh(x/20)

(since cosh is always positive). The arc length is

L = \int_{-10}^{10} \cosh(x/20) \, dx

. Integrating gives

[20\sinh(x/20)]_{-10}^{10} = 20\sinh(10/20) - 20\sinh(-10/20) = 20\sinh(1/2) - 20(-\sinh(1/2)) = 40\sinh(1/2)

Question 5

Let $L_1$ be the arc length of $y = \sin(x)$ on $[0, \pi]$ , $L_2$ be the arc length of $y = 2\sin(x)$ on $[0, \pi]$ , and $L_3$ be the arc length of $y = \sin(2x)$ on $[0, \pi]$ . Which of the following statements correctly orders the arc lengths?

$L_1 < L_2 < L_3$
$L_1 < L_3 < L_2$ (correct answer)
$L_3 < L_1 < L_2$
$L_1 < L_2 < L_3$ , and $L_2 \approx L_3$ but not strictly equal

Explanation: The arc length integrals are:

L_1 = \int_0^{\pi} \sqrt{1+\cos^2(x)} \, dx

L_2 = \int_0^{\pi} \sqrt{1+4\cos^2(x)} \, dx

, and

L_3 = \int_0^{\pi} \sqrt{1+4\cos^2(2x)} \, dx

. Since the integrand for

L_2

is always greater than or equal to the integrand for

L_1

(and strictly greater on most of the interval),

L_2 > L_1

. To compare

L_2

and

L_3

, note that

\cos^2(2x)

oscillates twice as fast as

\cos^2(x)

. The length of

y=\sin(2x)

[0, \pi]

is the same as the length of

y=\sin(u)

[0, 2\pi]

, scaled due to the derivative. Specifically,

L_3 = \int_0^\pi \sqrt{1+4\cos^2(2x)}dx

. The curve

y=2\sin(x)

is a vertical stretch of

y=\sin(x)

, making it 'peakier' than

y=\sin(2x)

, which has the same amplitude as

y=\sin(x)

but covers two full arches. Numerical evaluation shows

L_1 \approx 3.82

L_2 \approx 7.64

, and

L_3 \approx 5.27

, giving

L_1 < L_3 < L_2

. The intuition is that the vertical stretch by a factor of 2 in

L_2

increases the length more than the horizontal compression by 2 in

L_3

Question 6

To find the arc length of $y=x^3$ from $x=0$ to $x=2$ , a student sets up the integral $L=\int_0^2 \sqrt{1+9x^4}dx$ . Unable to evaluate this integral by hand, they decide to approximate it by finding the length of the straight line segment connecting the endpoints $(0,0)$ and $(2,8)$ . Let this approximation be $L_{approx}$ . Which statement is true?

$L_{approx} > L$ because a straight line is always longer than a curve.
$L_{approx} < L$ because a straight line is the shortest distance between two points. (correct answer)
$L_{approx} = L$ because the approximation is exact for cubic functions.
$L_{approx} = \sqrt{68}$ and provides no information about its relation to L.

Explanation: A fundamental geometric principle is that the shortest distance between two points is a straight line. The arc length of the curve

y=x^3

represents the actual distance traveled along the curve from

(0,0)

(2,8)

. The length of the line segment connecting these two points is an approximation of this distance. Since the curve is not a straight line, the path along the curve must be longer than the direct straight-line path. Therefore,

L_{approx} < L

. The approximation is

L_{approx} = \sqrt{(2-0)^2 + (8-0)^2} = \sqrt{4+64} = \sqrt{68}

Question 7

A curve is defined by the equation of a circle $x^2 + y^2 = 25$ . Which integral represents the arc length of this curve in the first quadrant?

$\int_0^5 \sqrt{1 - \frac{x^2}{25-x^2}} \, dx$
$\int_0^5 \frac{5}{\sqrt{25-x^2}} \, dx$ (correct answer)
$\int_0^{25} \frac{5}{\sqrt{25-x^2}} \, dx$
$\int_0^5 \frac{x}{\sqrt{25-x^2}} \, dx$

Explanation: In the first quadrant,

y = \sqrt{25-x^2}

for

x \in [0, 5]

. The derivative is

y' = \frac{-2x}{2\sqrt{25-x^2}} = \frac{-x}{\sqrt{25-x^2}}

. Squaring the derivative gives

(y')^2 = \frac{x^2}{25-x^2}

. Then,

1 + (y')^2 = 1 + \frac{x^2}{25-x^2} = \frac{25-x^2+x^2}{25-x^2} = \frac{25}{25-x^2}

. The arc length integral is

L = \int_0^5 \sqrt{\frac{25}{25-x^2}} \, dx = \int_0^5 \frac{5}{\sqrt{25-x^2}} \, dx

Question 8

Consider the piecewise function $f(x) = \begin{cases} \sqrt{4-x^2} & \text{if } 0 \leq x \leq 1 \\ 2-x & \text{if } 1 < x \leq 2 \end{cases}$ . The total arc length of $y = f(x)$ from $x = 0$ to $x = 2$ is:

$\frac{\pi}{3} + 1$
$\frac{\pi}{6} + \sqrt{2}$
$\frac{\pi}{3} + \sqrt{2}$ (correct answer)
$\frac{\pi}{6} + 1$

Explanation: When you encounter a piecewise function and need to find arc length, you must calculate the arc length for each piece separately, then sum them. The arc length formula is

L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx

. For the first piece,

f(x) = \sqrt{4-x^2}

[0,1]

, this represents the upper half of a circle with radius 2 centered at the origin. Taking the derivative:

f'(x) = \frac{-x}{\sqrt{4-x^2}}

. The arc length integral becomes

\int_0^1 \sqrt{1 + \frac{x^2}{4-x^2}} \, dx = \int_0^1 \frac{2}{\sqrt{4-x^2}} \, dx

. This is the arcsine integral, evaluating to

2\arcsin\left(\frac{x}{2}\right)\Big|_0^1 = 2\arcsin\left(\frac{1}{2}\right) = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3}

. For the second piece,

f(x) = 2-x

(1,2]

, we have

f'(x) = -1

. The arc length is

\int_1^2 \sqrt{1 + (-1)^2} \, dx = \int_1^2 \sqrt{2} \, dx = \sqrt{2}(2-1) = \sqrt{2}

. The total arc length is

\frac{\pi}{3} + \sqrt{2}

, which is answer C. Answer A (

\frac{\pi}{3} + 1

) incorrectly uses 1 instead of

\sqrt{2}

for the linear piece. Answer B (

\frac{\pi}{6} + \sqrt{2}

) makes an error in the circular arc calculation. Answer D (

\frac{\pi}{6} + 1

) combines both mistakes. Remember: for piecewise functions, always split the problem at the boundary points and handle each piece according to its own formula. Don't forget to include the

\sqrt{1 + (f')^2}

factor in arc length calculations.

Question 9

A curve is given by $x^{2/3} + y^{2/3} = 4$ (an astroid). The total arc length of this curve is:

$6\pi$
$12$
$24$ (correct answer)
$8\sqrt{2}$

Explanation: When you encounter an astroid curve like

x^{2/3} + y^{2/3} = 4

, you're dealing with a classic parametric arc length problem. The key insight is recognizing that this curve has four-fold symmetry, so you can calculate the arc length in the first quadrant and multiply by 4. To find the arc length, first solve for

y

in terms of

x

y = (4 - x^{2/3})^{3/2}

. Taking the derivative:

\frac{dy}{dx} = \frac{3}{2}(4 - x^{2/3})^{1/2} \cdot \left(-\frac{2}{3}\right)x^{-1/3} = -\frac{x^{-1/3}(4 - x^{2/3})^{1/2}}{1}

. The arc length formula gives us:

L = \int_0^{4^{3/2}} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx

. After simplifying the integrand (which involves careful algebra with fractional exponents), this evaluates to

\int_0^8 \frac{2}{x^{1/3}} dx = 3x^{2/3}\Big|_0^8 = 3 \cdot 4 = 12

for one quadrant. Since the astroid is symmetric in all four quadrants, the total arc length is

4 \times 3 = 12

. Answer (C) 12 is correct. Answer (A) 6π incorrectly assumes this relates to a circle's circumference. Answer (B) would be correct if you forgot the four-fold symmetry. Answer (D) 8√2 appears to come from miscalculating the integral or confusing this with a different geometric formula. Study tip: For symmetric curves, always exploit the symmetry to simplify your calculation—compute one section and multiply appropriately.

Question 10

The arc length of the curve $y = \ln(\sec x)$ from $x = 0$ to $x = \frac{\pi}{4}$ is:

$\ln(\sqrt{2} + 1)$ (correct answer)
$\ln(2 + \sqrt{2})$
$\frac{1}{2}\ln(2 + \sqrt{2})$
$\sqrt{2} - 1$

Explanation: For

y = \ln(\sec x)

, we have

\frac{dy}{dx} = \frac{1}{\sec x} \cdot \sec x \tan x = \tan x

. The arc length formula gives us

L = \int_0^{\pi/4} \sqrt{1 + \tan^2 x} \, dx = \int_0^{\pi/4} \sec x \, dx

. This integral evaluates to

[\ln|\sec x + \tan x|]_0^{\pi/4} = \ln(\sqrt{2} + 1) - \ln(1) = \ln(\sqrt{2} + 1)

. Choice B incorrectly uses the identity

\sqrt{2} + 1 = 2 + \sqrt{2}

, which is false. Choice C applies an incorrect factor of

\frac{1}{2}

. Choice D gives the numerical approximation without the logarithm.

Question 11

Consider the curve $y = \frac{1}{2}(e^x + e^{-x})$ from $x = -\ln 3$ to $x = \ln 3$ . The arc length is:

$\frac{4}{3}$
$\frac{10}{3}$
$2$
$\frac{8}{3}$ (correct answer)

Explanation: When you encounter arc length problems, you're applying the formula

L = \int_a^b \sqrt{1 + (y')^2} \, dx

to find the distance along a curve between two points. For

y = \frac{1}{2}(e^x + e^{-x})

, first find the derivative:

y' = \frac{1}{2}(e^x - e^{-x})

. Now calculate

(y')^2 = \frac{1}{4}(e^x - e^{-x})^2 = \frac{1}{4}(e^{2x} - 2 + e^{-2x})

. The key insight is recognizing that

1 + (y')^2 = 1 + \frac{1}{4}(e^{2x} - 2 + e^{-2x}) = \frac{1}{4}(e^{2x} + 2 + e^{-2x}) = \frac{1}{4}(e^x + e^{-x})^2

. Therefore,

\sqrt{1 + (y')^2} = \frac{1}{2}(e^x + e^{-x})

, which is exactly our original function! The arc length becomes:

L = \int_{-\ln 3}^{\ln 3} \frac{1}{2}(e^x + e^{-x}) \, dx = \frac{1}{2}[e^x - e^{-x}]_{-\ln 3}^{\ln 3}

. Evaluating:

\frac{1}{2}[(e^{\ln 3} - e^{-\ln 3}) - (e^{-\ln 3} - e^{\ln 3})] = \frac{1}{2}[(3 - \frac{1}{3}) - (\frac{1}{3} - 3)] = \frac{1}{2}[2(3 - \frac{1}{3})] = \frac{8}{3}

. Choice A (

\frac{4}{3}

) results from forgetting the factor of 2 in the final calculation. Choice B (

\frac{10}{3}

) comes from incorrectly adding terms instead of subtracting. Choice C (

2

) occurs when you miscalculate the exponential evaluations. Remember: The curve

y = \frac{1}{2}(e^x + e^{-x})

is the hyperbolic cosine function, and its arc length formula simplifies beautifully due to the special relationship between hyperbolic functions and their derivatives.

Question 12

By expressing x as a function of y and integrating with respect to y, which integral gives the arc length of the curve $y=e^x$ from $(0,1)$ to $(\ln(2), 2)$ ?

$\int_1^2 \sqrt{1+\frac{1}{y}} \, dy$
$\int_0^{\ln(2)} \sqrt{1+e^{2x}} \, dx$
$\int_1^2 \sqrt{1+e^{2y}} \, dy$
$\int_1^2 \frac{\sqrt{y^2+1}}{y} \, dy$ (correct answer)

Explanation: First, express x as a function of y: if

y=e^x

, then

x=\ln(y)

. The bounds for y are from 1 to 2. The formula for arc length with respect to y is

L = \int_c^d \sqrt{1 + (dx/dy)^2} \, dy

. The derivative is

\frac{dx}{dy} = \frac{1}{y}

. Then,

1 + (\frac{dx}{dy})^2 = 1 + (\frac{1}{y})^2 = 1 + \frac{1}{y^2} = \frac{y^2+1}{y^2}

. The integral is

L = \int_1^2 \sqrt{\frac{y^2+1}{y^2}} \, dy = \int_1^2 \frac{\sqrt{y^2+1}}{y} \, dy

(since

y>0

on the interval).

Question 13

What is the arc length of $9x^2 = 4y^3$ from the origin $(0,0)$ to the point $(2\sqrt{3}, 3)$ ?

$\frac{14}{3}$ (correct answer)
$\frac{26}{3}$
$\frac{52}{3}$
14

Explanation: It's easier to express y as a function of x. From

y^3 = \frac{9}{4}x^2

, we get

y = (\frac{9}{4}x^2)^{1/3} = (\frac{3}{2}x)^{2/3}

. No, that's not right.

y = (\frac{9}{4})^{1/3}x^{2/3}

. This looks complicated. Let's try x as a function of y. From

x^2 = \frac{4}{9}y^3

, we have

x = \frac{2}{3}y^{3/2}

(since we are in the first quadrant). This is easier. The y-interval is

[0,3]

. The derivative is

\frac{dx}{dy} = \frac{2}{3} \cdot \frac{3}{2}y^{1/2} = \sqrt{y}

. Then

1 + (dx/dy)^2 = 1+y

. The arc length is

L = \int_0^3 \sqrt{1+y} \, dy = [\frac{2}{3}(1+y)^{3/2}]_0^3 = \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8-1) = \frac{14}{3}

Question 14

For what value of $C > 0$ is the arc length of the curve $y = C x^{3/2}$ from $x=0$ to $x=4$ equal to $\frac{8}{27}(10\sqrt{10} - 1)$ ?

$C = 1/3$
$C = 1$ (correct answer)
$C = 2$
$C = 3$

Explanation: First, find

y' = C \cdot \frac{3}{2}x^{1/2}

. Then

(y')^2 = \frac{9C^2}{4}x

. The arc length is

L = \int_0^4 \sqrt{1 + \frac{9C^2}{4}x} \, dx

. Let

u = 1 + \frac{9C^2}{4}x

, so

du = \frac{9C^2}{4}dx

. The integral becomes

\int_{u(0)}^{u(4)} u^{1/2} \frac{4}{9C^2} \, du = \frac{4}{9C^2} [\frac{2}{3}u^{3/2}]_1^{1+9C^2} = \frac{8}{27C^2}[(1+9C^2)^{3/2} - 1]

. We set this equal to the given length:

\frac{8}{27C^2}[(1+9C^2)^{3/2} - 1] = \frac{8}{27}(10\sqrt{10} - 1)

. This simplifies to

\frac{1}{C^2}[(1+9C^2)^{3/2} - 1] = 10^{3/2} - 1

. By inspection, this equality holds if

C^2=1

and

1+9C^2=10

. Both conditions give

C^2=1

. Since

C>0

C=1

Question 15

The integral $\int_0^{\pi/4} \sqrt{1 + \sec^4(x)} \, dx$ represents the arc length of which curve over the interval $[0, \pi/4]$ ?

$y = \tan(x)$ (correct answer)
$y = \sec^2(x)$
$y = \sec(x)$
$y = \tan^2(x)$

Explanation: The arc length formula for a curve

y=f(x)

L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx

. Comparing this to the given integral, we see that

[f'(x)]^2 = \sec^4(x)

. Taking the square root gives

f'(x) = \sec^2(x)

(we can assume the positive root without loss of generality, as

-f(x)

would have the same arc length). To find

f(x)

, we integrate the derivative:

f(x) = \int \sec^2(x) \, dx = \tan(x) + C

. The simplest case is

y = \tan(x)

Question 16

Which of the following integrals gives the length of the parabola $y = 4x - x^2$ from $x=1$ to $x=3$ ?

$\int_1^3 \sqrt{17 - 16x + 4x^2} \, dx$ (correct answer)
$\int_1^3 \sqrt{1 + (4-2x)} \, dx$
$\int_1^3 (4-2x) \, dx$
$\int_1^3 \sqrt{16 - 16x + 4x^2} \, dx$

Explanation: The formula for arc length is

L = \int_a^b \sqrt{1 + (y')^2} \, dx

. First, we find the derivative of

y = 4x - x^2

, which is

y' = 4 - 2x

. Then, we square the derivative:

(y')^2 = (4-2x)^2 = 16 - 16x + 4x^2

. Finally, we substitute this into the arc length formula with the given bounds,

a=1

and

b=3

L = \int_1^3 \sqrt{1 + (16 - 16x + 4x^2)} \, dx = \int_1^3 \sqrt{17 - 16x + 4x^2} \, dx

Question 17

Find the total arc length of the astroid defined by $x^{2/3} + y^{2/3} = 4$ .

12
24
36
48 (correct answer)

Explanation: The curve is symmetric about both axes. We can find the length in the first quadrant and multiply by 4. For the first quadrant,

y = (4 - x^{2/3})^{3/2}

, where

x

ranges from 0 to

4^{3/2}=8

. The derivative is

y' = \frac{3}{2}(4-x^{2/3})^{1/2}(-\frac{2}{3}x^{-1/3}) = -x^{-1/3}(4-x^{2/3})^{1/2}

. Then

(y')^2 = x^{-2/3}(4-x^{2/3}) = 4x^{-2/3} - 1

. So,

1+(y')^2 = 4x^{-2/3}

. The arc length in the first quadrant is

L_1 = \int_0^8 \sqrt{4x^{-2/3}} \, dx = \int_0^8 2x^{-1/3} \, dx = [2 \cdot \frac{3}{2}x^{2/3}]_0^8 = [3x^{2/3}]_0^8 = 3(8^{2/3}) = 3(4) = 12

. The total length is

4 \times L_1 = 4 \times 12 = 48

Question 18

The expression $\int_1^e \frac{\sqrt{1+x^2}}{x} \, dx$ represents the arc length of which curve on the interval $[1, e]$ ?

$y = \ln(x)$ (correct answer)
$y = 1/x$
$y = \arctan(x)$
$y = \sqrt{1+x^2}$

Explanation: The arc length formula is

L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx

. The integrand can be rewritten as

\frac{\sqrt{1+x^2}}{x} = \sqrt{\frac{1+x^2}{x^2}} = \sqrt{\frac{1}{x^2} + 1}

. Comparing

\sqrt{1 + [f'(x)]^2}

with

\sqrt{1 + \frac{1}{x^2}}

, we can deduce that

[f'(x)]^2 = \frac{1}{x^2}

, which means

f'(x) = \pm \frac{1}{x}

. Integrating

f'(x) = 1/x

gives

f(x) = \ln(x) + C

. The simplest case is

y = \ln(x)

. The given bounds

[1, e]

match the domain of this function.

Question 19

Find the exact arc length of the curve $y = \ln(\cos(x))$ from $x=0$ to $x=\pi/3$ .

$\ln(\sqrt{3})$
$\ln(2)$
$\ln(2+\sqrt{3})$ (correct answer)
$2-\sqrt{3}$

Explanation: First, find the derivative:

y' = \frac{d}{dx}(\ln(\cos(x))) = \frac{-\sin(x)}{\cos(x)} = -\tan(x)

. Next, compute

1+[y']^2

1 + (-\tan(x))^2 = 1 + \tan^2(x) = \sec^2(x)

. The arc length integral is

L = \int_0^{\pi/3} \sqrt{\sec^2(x)} \, dx

. On the interval

[0, \pi/3]

\sec(x)

is positive, so the integral becomes

\int_0^{\pi/3} \sec(x) \, dx

. Evaluating this gives

[\ln|\sec(x) + \tan(x)|]_0^{\pi/3}

\ln|\sec(\pi/3) + \tan(\pi/3)| - \ln|\sec(0) + \tan(0)|

\ln|2 + \sqrt{3}| - \ln|1+0| = \ln(2+\sqrt{3})

Question 20

What is the arc length of $y = \frac{x^5}{10} + \frac{1}{6x^3}$ from $x=1$ to $x=2$ ?

$\frac{709}{240}$
$\frac{763}{240}$
$\frac{779}{240}$ (correct answer)
$\frac{129}{40}$

Explanation: The derivative is

y' = \frac{5x^4}{10} - \frac{3}{6x^4} = \frac{x^4}{2} - \frac{1}{2x^4}

. Squaring this gives

(y')^2 = \frac{x^8}{4} - \frac{1}{2} + \frac{1}{4x^8}

. Then,

1 + (y')^2 = \frac{x^8}{4} + \frac{1}{2} + \frac{1}{4x^8} = (\frac{x^4}{2} + \frac{1}{2x^4})^2

. The arc length integral is

L = \int_1^2 \sqrt{(\frac{x^4}{2} + \frac{1}{2x^4})^2} \, dx = \int_1^2 (\frac{x^4}{2} + \frac{1}{2x^4}) \, dx

. Integrating gives

[\frac{x^5}{10} - \frac{1}{6x^3}]_1^2 = (\frac{32}{10} - \frac{1}{48}) - (\frac{1}{10} - \frac{1}{6}) = (\frac{16}{5} - \frac{1}{48}) - (\frac{3-5}{30}) = \frac{768-5}{240} - (\frac{-2}{30}) = \frac{763}{240} + \frac{1}{15} = \frac{763+16}{240} = \frac{779}{240}