Arc Length (Cartesian) - Calculus 2

1. What is the arc length of $9x^2 = 4y^3$ from the origin $(0,0)$ to the point $(2\sqrt{3}, 3)$?
   Answer: $\frac{14}{3}$ 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 $L = \int_0^3 \sqrt{1+y} \, dy = [\frac{2}{3}(1+y)^{3/2}]_0^3 = \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8-1) = \frac{14}{3}$.
2. 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?
   Answer: $L_1 < L_3 < L_2$ 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 $y=\sin(2x)$ on $[0, \pi]$ is the same as the length of $y=\sin(u)$ on $[0, 2\pi]$, 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 $L_3$.
3. Consider the curve $y = \frac{1}{2}(e^x + e^{-x})$ from $x = -\ln 3$ to $x = \ln 3$. The arc length is:
   Answer: $\frac{8}{3}$ 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.
4. Find the exact arc length of the curve $y = \ln(\cos(x))$ from $x=0$ to $x=\pi/3$.
   Answer: $\ln(2+\sqrt{3})$ 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 $[\ln|\sec(x) + \tan(x)|]_0^{\pi/3}$ = $\ln|\sec(\pi/3) + \tan(\pi/3)| - \ln|\sec(0) + \tan(0)|$ = $\ln|2 + \sqrt{3}| - \ln|1+0| = \ln(2+\sqrt{3})$.
5. 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?
   Answer: $40\sinh(1/2)$ 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 $[20\sinh(x/20)]_{-10}^{10} = 20\sinh(10/20) - 20\sinh(-10/20) = 20\sinh(1/2) - 20(-\sinh(1/2)) = 40\sinh(1/2)$.
6. A curve is defined by $x = t^3 - 3t$ and $y = t^2 + 1$ for $t \in [-2, 2]$. If the curve crosses itself at exactly one point, what is the arc length of the entire curve?
   Answer: $\int_{-2}^2 \sqrt{(3t^2 - 3)^2 + (2t)^2} \, dt$ For parametric curves, arc length is $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt$. Here, $\frac{dx}{dt} = 3t^2 - 3$ and $\frac{dy}{dt} = 2t$. The fact that the curve crosses itself doesn't affect the arc length calculation - we still integrate over the full parameter range. Thus $L = \int_{-2}^2 \sqrt{(3t^2 - 3)^2 + (2t)^2} \, dt$. Choice A incorrectly factors and uses symmetry where none exists. Choice C expands incorrectly: $(3t^2-3)^2 = 9t^4 - 18t^2 + 9$, not $9t^4 - 18t^2 + 13$. Choice D incorrectly attempts to use the Cartesian formula.
7. 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 $(\ln(2), 2)$?
   Answer: $\int_1^2 \frac{\sqrt{y^2+1}}{y} \, dy$ 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).
8. Find the total arc length of the astroid defined by $x^{2/3} + y^{2/3} = 4$.
   Answer: 48 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 $4 \times L_1 = 4 \times 12 = 48$.
9. 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?
   Answer: $L_{approx} < L$ because a straight line is the shortest distance between two points. 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 $(0,0)$ to $(2,8)$. 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 $L_{approx} = \sqrt{(2-0)^2 + (8-0)^2} = \sqrt{4+64} = \sqrt{68}$.
10. The arc length of the curve $y = \ln(\sec x)$ from $x = 0$ to $x = \frac{\pi}{4}$ is:
    Answer: $\ln(\sqrt{2} + 1)$ 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.
11. Which of the following integrals gives the length of the parabola $y = 4x - x^2$ from $x=1$ to $x=3$?
    Answer: $\int_1^3 \sqrt{17 - 16x + 4x^2} \, dx$ 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 $b=3$: $L = \int_1^3 \sqrt{1 + (16 - 16x + 4x^2)} \, dx = \int_1^3 \sqrt{17 - 16x + 4x^2} \, dx$.
12. The arc length of $y = \frac{2}{3}x^{3/2}$ from $x=0$ to $x=a$ is $\frac{52}{3}$. What is the value of $a$?
    Answer: $a = 8$ First, find the derivative: $y' = \frac{2}{3} \cdot \frac{3}{2}x^{1/2} = \sqrt{x}$. Then $1+(y')^2 = 1+x$. The arc length is $L = \int_0^a \sqrt{1+x} \, dx$. Let $u=1+x$, $du=dx$. The integral becomes $\int_1^{1+a} u^{1/2} \, du = [\frac{2}{3}u^{3/2}]_1^{1+a} = \frac{2}{3}((1+a)^{3/2} - 1)$. Setting this equal to $\frac{52}{3}$: $\frac{2}{3}((1+a)^{3/2} - 1) = \frac{52}{3}$. This gives $(1+a)^{3/2} - 1 = 26$, so $(1+a)^{3/2} = 27$. Therefore $1+a = 27^{2/3} = 9$, and $a = 8$.
13. 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 $L_2$?
    Answer: $L_1 = L_2$ 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 $(0,0)$ to $(1,1)$ 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.
14. The arc length of a curve is given by the integral $L = \int_0^2 \sqrt{1 + 16x^6} \, dx$. If the curve passes through the origin $(0,0)$, which of the following could be the equation of the curve?
    Answer: $y = x^4$ The arc length formula is $L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx$. Comparing this with the given integral, we can identify $[f'(x)]^2 = 16x^6$. Taking the square root gives $f'(x) = 4x^3$ (or $-4x^3$, which would result in the same length). To find $f(x)$, we integrate the derivative: $f(x) = \int 4x^3 \, dx = x^4 + C$. Since the curve passes through the origin $(0,0)$, we have $0 = 0^4 + C$, which implies $C=0$. Therefore, the equation of the curve is $y = x^4$.
15. 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?
    Answer: $\int_0^5 \frac{5}{\sqrt{25-x^2}} \, dx$ 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 $L = \int_0^5 \sqrt{\frac{25}{25-x^2}} \, dx = \int_0^5 \frac{5}{\sqrt{25-x^2}} \, dx$.
16. 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?
    Answer: $L_2 > L_1$ 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 $x$ in $(0,1]$, 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 $L_1$.
17. What is the arc length of $y = \frac{x^5}{10} + \frac{1}{6x^3}$ from $x=1$ to $x=2$?
    Answer: $\frac{779}{240}$ 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 $[\frac{x^5}{10} - \frac{1}{6x^3}]_1^2 = (\frac{32}{10} - \frac{1}{48}) - (\frac{1}{10} - \frac{1}{6}) = (\frac{16}{5} - \frac{1}{48}) - (\frac{3-5}{30}) = \frac{768-5}{240} - (\frac{-2}{30}) = \frac{763}{240} + \frac{1}{15} = \frac{763+16}{240} = \frac{779}{240}$.
18. The arc length of the curve $y = \int_1^x \sqrt{t^4 - 1} \, dt$ from $x = 1$ to $x = 2$ is:
    Answer: $\int_1^2 x^2 \, dx$ When you encounter arc length problems involving parametrically defined or implicitly defined curves, remember that the arc length formula requires finding $\frac{dy}{dx}$ first, then applying $L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$. Since $y = \int_1^x \sqrt{t^4 - 1} \, dt$, you need to find $\frac{dy}{dx}$. By the Fundamental Theorem of Calculus, when you differentiate an integral with respect to its upper limit, you get the integrand evaluated at that limit. Therefore: $\frac{dy}{dx} = \sqrt{x^4 - 1}$. Now apply the arc length formula: $L = \int_1^2 \sqrt{1 + \left(\sqrt{x^4 - 1}\right)^2} \, dx = \int_1^2 \sqrt{1 + x^4 - 1} \, dx = \int_1^2 \sqrt{x^4} \, dx = \int_1^2 x^2 \, dx$ This confirms answer D is correct. A stops at $\sqrt{1 + x^4 - 1}$ without simplifying to $\sqrt{x^4} = x^2$. B incorrectly computes $\left(\frac{dy}{dx}\right)^2$ as $2x^4 - x^8$, suggesting a fundamental error in squaring $\sqrt{x^4 - 1}$. C appears to expand $(\sqrt{x^4 - 1})^2$ incorrectly as $(x^2 - 1)^2$, which would give $x^4 - 2x^2 + 1$, not $x^4 - 1$. Key strategy: When finding arc length of curves defined by integrals, always use the Fundamental Theorem of Calculus to find $\frac{dy}{dx}$ first, then simplify your expression completely before concluding.
19. 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:
    Answer: $\frac{\pi}{3} + \sqrt{2}$ 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, $f(x) = \sqrt{4-x^2}$ on $[0,1]$, 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, $f(x) = 2-x$ on $(1,2]$, 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.
20. The expression $\int_1^e \frac{\sqrt{1+x^2}}{x} \, dx$ represents the arc length of which curve on the interval $[1, e]$?
    Answer: $y = \ln(x)$ 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.

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