Arc Length & Distance Traveled - Calculus 2

1. A particle's position is given by $x(t) = \int_0^t \cos(u^2) \, du$ and $y(t) = \int_0^t \sin(u^2) \, du$. What is the total distance traveled by the particle from $t=0$ to $t=5$?
   Answer: 5 To find the distance traveled, we need to integrate the particle's speed. The speed is $s(t) = \sqrt{(x'(t))^2 + (y'(t))^2}$. Using the Fundamental Theorem of Calculus Part 1, we can find the derivatives of the position functions: $x'(t) = \cos(t^2)$ and $y'(t) = \sin(t^2)$. Now, we calculate the speed: $s(t) = \sqrt{(\cos(t^2))^2 + (\sin(t^2))^2} = \sqrt{\cos^2(t^2) + \sin^2(t^2)}$. Using the trigonometric identity $\cos^2(\theta) + \sin^2(\theta) = 1$, the speed simplifies to $s(t) = \sqrt{1} = 1$. The particle travels at a constant speed of 1. The total distance traveled from $t=0$ to $t=5$ is $\int_0^5 1 \, dt = [t]_0^5 = 5$.
2. A wire is shaped like the curve $y = \cosh(x)$ from $x = -\ln(2)$ to $x = \ln(2)$. What is the length of the wire?
   Answer: $\frac{3}{2}$ The length of the wire is the arc length of the curve $y=\cosh(x)$. The derivative is $y' = \sinh(x)$. The arc length integrand is $\sqrt{1+(y')^2} = \sqrt{1+\sinh^2(x)}$. Using the identity $\cosh^2(x) - \sinh^2(x) = 1$, we have $1+\sinh^2(x) = \cosh^2(x)$. So the integrand is $\sqrt{\cosh^2(x)} = \cosh(x)$ (since $\cosh(x)$ is always positive). The length is $L = \int_{-\ln(2)}^{\ln(2)} \cosh(x) \, dx = [\sinh(x)]_{-\ln(2)}^{\ln(2)} = \sinh(\ln(2)) - \sinh(-\ln(2))$. Since $\sinh(x)$ is an odd function, $\sinh(-\ln(2)) = -\sinh(\ln(2))$, so the length is $2\sinh(\ln(2))$. Using the definition $\sinh(z) = (e^z - e^{-z})/2$, we have $\sinh(\ln(2)) = (e^{\ln(2)} - e^{-\ln(2)})/2 = (2 - 1/2)/2 = (3/2)/2 = 3/4$. The total length is $2 \cdot (3/4) = 3/2$.
3. Consider the curve given by $y = x^2$ for $0 \le x \le 1$. The arc length is $L_1 = \int_0^1 \sqrt{1+4x^2}dx$. Now consider the curve $x=y^2$ for $0 \le y \le 1$. Its arc length is $L_2 = \int_0^1 \sqrt{1+4y^2}dy$. Which of the following statements is true?
   Answer: $L_1 = L_2$ The arc length for $y=x^2$ from $x=0$ to $x=1$ is given by $L_1 = \int_0^1 \sqrt{1+(2x)^2}dx = \int_0^1 \sqrt{1+4x^2}dx$. The arc length for $x=y^2$ from $y=0$ to $y=1$ is given by $L_2 = \int_0^1 \sqrt{1+(2y)^2}dy = \int_0^1 \sqrt{1+4y^2}dy$. The two integrals are definite integrals of the same function, $f(z) = \sqrt{1+4z^2}$, over the same interval $[0,1]$. The variable of integration (whether it's called $x$ or $y$) is a dummy variable and does not affect the value of the definite integral. Therefore, the values of the two integrals are equal, and $L_1 = L_2$. Geometrically, the curve $x=y^2$ for $0 \le y \le 1$ is a reflection of the curve $y=x^2$ for $0 \le x \le 1$ across the line $y=x$, and reflection is a rigid transformation that preserves arc length.
4. A particle moves along a path described by $x(t) = e^t \cos(t)$, $y(t) = e^t \sin(t)$ for $t \in [0, \pi]$. The particle's speed varies according to its position. What is the total distance traveled?
   Answer: $\sqrt{2}(e^{\pi} - 1)$ The distance traveled is $\int_0^{\pi} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. We have $\frac{dx}{dt} = e^t \cos(t) - e^t \sin(t) = e^t(\cos(t) - \sin(t))$ and $\frac{dy}{dt} = e^t \sin(t) + e^t \cos(t) = e^t(\sin(t) + \cos(t))$. Therefore $(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 = e^{2t}[(\cos(t) - \sin(t))^2 + (\sin(t) + \cos(t))^2] = e^{2t}[\cos^2(t) - 2\cos(t)\sin(t) + \sin^2(t) + \sin^2(t) + 2\sin(t)\cos(t) + \cos^2(t)] = e^{2t}[2\cos^2(t) + 2\sin^2(t)] = 2e^{2t}$. So the distance is $\int_0^{\pi} \sqrt{2e^{2t}} dt = \sqrt{2} \int_0^{\pi} e^t dt = \sqrt{2}[e^t]_0^{\pi} = \sqrt{2}(e^{\pi} - 1)$. Choice B omits the $\sqrt{2}$ factor. Choice C forgets to subtract 1. Choice D incorrectly computes the coefficient.
5. Which expression represents the arc length of the curve $y = x|x|$ from $x = -2$ to $x = 2$?
   Answer: $\int_{-2}^2 \sqrt{1 + 4x^2} \, dx$ The function can be written piecewise: $y = x^2$ for $x \ge 0$ and $y = -x^2$ for $x < 0$. The derivatives are $y' = 2x$ for $x > 0$ and $y' = -2x$ for $x < 0$. The arc length integrand is $\sqrt{1+(y')^2}$. For $x > 0$, this is $\sqrt{1+(2x)^2} = \sqrt{1+4x^2}$. For $x < 0$, this is $\sqrt{1+(-2x)^2} = \sqrt{1+4x^2}$. Since the integrand is the same for both pieces, the total arc length is the sum of the integrals over the two parts: $\int_{-2}^0 \sqrt{1+4x^2} \, dx + \int_0^2 \sqrt{1+4x^2} \, dx$. This can be combined into a single integral: $\int_{-2}^2 \sqrt{1+4x^2} \, dx$.
6. Which integral represents the distance traveled by a particle from $t=0$ to $t=2$ whose position is given by $x(t) = t^2$ and $y(t) = t^3 - 3t$?
   Answer: $\int_0^2 \sqrt{4t^2 + (3t^2-3)^2} \, dt$ The distance traveled is the integral of the speed, which is $s(t) = \sqrt{(x'(t))^2 + (y'(t))^2}$. First, we find the derivatives of the position functions: $x'(t) = 2t$ and $y'(t) = 3t^2 - 3$. Next, we square these derivatives: $(x'(t))^2 = (2t)^2 = 4t^2$ and $(y'(t))^2 = (3t^2 - 3)^2$. The speed is $s(t) = \sqrt{4t^2 + (3t^2-3)^2}$. The total distance traveled is the integral of the speed over the given interval, which is $\int_0^2 \sqrt{4t^2 + (3t^2-3)^2} \, dt$. Note that the expression under the radical simplifies to $\sqrt{9t^4 - 14t^2 + 9}$, which is not a perfect square, so the question only asks for the setup.
7. The velocity of a particle moving in the xy-plane is $\vec{v}(t) = \langle \sin(t), t \cos(t) \rangle$. Which integral represents the total distance traveled from $t=0$ to $t=\pi$?
   Answer: $\int_0^{\pi} \sqrt{\sin^2(t) + t^2 \cos^2(t)} \, dt$ The distance traveled is the integral of the speed. The speed is the magnitude of the velocity vector: $s(t) = ||\vec{v}(t)|| = \sqrt{(v_x(t))^2 + (v_y(t))^2}$. The components of the velocity are given as $v_x(t) = \sin(t)$ and $v_y(t) = t \cos(t)$. Squaring and adding these components gives $(\sin(t))^2 + (t \cos(t))^2 = \sin^2(t) + t^2 \cos^2(t)$. The speed is $s(t) = \sqrt{\sin^2(t) + t^2 \cos^2(t)}$. Integrating the speed from $t=0$ to $t=\pi$ gives the total distance traveled. Therefore, the correct integral is $\int_0^{\pi} \sqrt{\sin^2(t) + t^2 \cos^2(t)} \, dt$.
8. Find the arc length of the polar curve $r = e^{2\theta}$ from $\theta=0$ to $\theta=\ln(2)$.
   Answer: $\frac{3\sqrt{5}}{2}$ The arc length formula for a polar curve is $L = \int_a^b \sqrt{r^2 + (dr/d\theta)^2} \, d\theta$. Here, $r = e^{2\theta}$, so $dr/d\theta = 2e^{2\theta}$. The term under the square root is $r^2 + (dr/d\theta)^2 = (e^{2\theta})^2 + (2e^{2\theta})^2 = e^{4\theta} + 4e^{4\theta} = 5e^{4\theta}$. The integrand is $\sqrt{5e^{4\theta}} = \sqrt{5}e^{2\theta}$. The integral for the arc length is $L = \int_0^{\ln(2)} \sqrt{5}e^{2\theta} \, d\theta = \sqrt{5}[\frac{1}{2}e^{2\theta}]_0^{\ln(2)}$. Evaluating this gives $\frac{\sqrt{5}}{2}(e^{2\ln(2)} - e^0) = \frac{\sqrt{5}}{2}(e^{\ln(4)} - 1) = \frac{\sqrt{5}}{2}(4-1) = \frac{3\sqrt{5}}{2}$.
9. The arc length of the curve $y = \ln(\cos x)$ from $x=0$ to $x=k$, where $0 < k < \pi/2$, is found to be $\ln(2+\sqrt{3})$. What is the value of $k$?
   Answer: $\pi/3$ First, we set up the arc length integral. The derivative is $y' = \frac{1}{\cos x}(-\sin x) = -\tan x$. The arc length integrand is $\sqrt{1+(y')^2} = \sqrt{1+\tan^2 x} = \sqrt{\sec^2 x} = |\sec x|$. Since $0 < k < \pi/2$, $\sec x$ is positive, so the integrand is $\sec x$. The arc length is $L = \int_0^k \sec x \, dx = [\ln|\sec x + \tan x|]_0^k$. Evaluating the integral gives $\ln|\sec k + \tan k| - \ln|\sec 0 + \tan 0| = \ln(\sec k + \tan k) - \ln(1+0) = \ln(\sec k + \tan k)$. We are given that this length is $\ln(2+\sqrt{3})$. Therefore, $\sec k + \tan k = 2+\sqrt{3}$. We can test the given angle values. For $k=\pi/3$, $\sec(\pi/3) = 2$ and $\tan(\pi/3) = \sqrt{3}$. Their sum is $2+\sqrt{3}$, so $k=\pi/3$.
10. The astroid curve is given parametrically by $x = a\cos^3(t)$, $y = a\sin^3(t)$ where $a > 0$ and $t \in [0, 2\pi]$. If the total arc length of the astroid is $12$, what is the value of $a$?
    Answer: $2$ For the astroid, $\frac{dx}{dt} = -3a\cos^2(t)\sin(t)$ and $\frac{dy}{dt} = 3a\sin^2(t)\cos(t)$. So $(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 = 9a^2\cos^4(t)\sin^2(t) + 9a^2\sin^4(t)\cos^2(t) = 9a^2\cos^2(t)\sin^2(t)[\cos^2(t) + \sin^2(t)] = 9a^2\cos^2(t)\sin^2(t)$. The arc length for one complete cycle is $L = \int_0^{2\pi} 3a|\cos(t)\sin(t)| dt$. Due to symmetry, we can compute $4\int_0^{\pi/2} 3a\cos(t)\sin(t) dt = 4 \cdot 3a \int_0^{\pi/2} \frac{1}{2}\sin(2t) dt = 6a[-\frac{1}{2}\cos(2t)]_0^{\pi/2} = 6a[-\frac{1}{2}(-1 - 1)] = 6a$. Setting $6a = 12$ gives $a = 2$. Choice A gives $a = \frac{4}{3}$, which would yield arc length $8$. Choice C gives $a = \frac{8}{3}$, yielding arc length $16$. Choice D gives $a = 3$, yielding arc length $18$.
11. Which of the following integrals correctly represents the arc length of the astroid $x^{2/3} + y^{2/3} = a^{2/3}$ in the first quadrant?
    Answer: $\int_0^a a^{1/3}x^{-1/3} \, dx$ To find the arc length, we first need $dy/dx$. Using implicit differentiation on $x^{2/3} + y^{2/3} = a^{2/3}$: $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0$, which gives $\frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}}$. The arc length integrand requires $1+(dy/dx)^2$. So, $1 + (-\frac{y^{1/3}}{x^{1/3}})^2 = 1 + \frac{y^{2/3}}{x^{2/3}} = \frac{x^{2/3} + y^{2/3}}{x^{2/3}}$. From the original equation, $x^{2/3} + y^{2/3} = a^{2/3}$, so this simplifies to $\frac{a^{2/3}}{x^{2/3}}$. The arc length element is $ds = \sqrt{\frac{a^{2/3}}{x^{2/3}}} \, dx = \frac{a^{1/3}}{x^{1/3}} \, dx = a^{1/3}x^{-1/3} \, dx$. In the first quadrant, $x$ ranges from 0 to $a$. So the integral is $\int_0^a a^{1/3}x^{-1/3} \, dx$.
12. The length of the curve $y = \int_1^x \sqrt{t^2-1} \, dt$ from $x=1$ to $x=3$ is requested. Which of the following statements is correct?
    Answer: The arc length is 4. To find the arc length, we need the derivative $y'$. By the Fundamental Theorem of Calculus Part 1, if $y = \int_1^x \sqrt{t^2-1} \, dt$, then $y' = \frac{dy}{dx} = \sqrt{x^2-1}$. The arc length formula is $L = \int_1^3 \sqrt{1+(y')^2} \, dx$. Substituting $y'$, we get $1+(y')^2 = 1 + (\sqrt{x^2-1})^2 = 1 + x^2 - 1 = x^2$. The integral becomes $L = \int_1^3 \sqrt{x^2} \, dx$. Since the interval is $[1, 3]$, $x$ is positive, so $\sqrt{x^2}=x$. The integral is $L = \int_1^3 x \, dx = [\frac{x^2}{2}]_1^3 = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4$.
13. The arc length of the curve $y = \frac{2}{3}x^{3/2}$ from $x=0$ to $x=b$ is $\frac{14}{3}$. What is the value of $b$?
    Answer: 3 First, find the derivative: $y' = \frac{2}{3} \cdot \frac{3}{2} x^{1/2} = \sqrt{x}$. Then, set up the arc length integral: $L = \int_0^b \sqrt{1 + (y')^2} \, dx = \int_0^b \sqrt{1 + (\sqrt{x})^2} \, dx = \int_0^b \sqrt{1+x} \, dx$. We evaluate this integral: $\int_0^b (1+x)^{1/2} \, dx = [\frac{2}{3}(1+x)^{3/2}]_0^b = \frac{2}{3}(1+b)^{3/2} - \frac{2}{3}(1+0)^{3/2} = \frac{2}{3}((1+b)^{3/2} - 1)$. We are given that the length is $\frac{14}{3}$. So, $\frac{2}{3}((1+b)^{3/2} - 1) = \frac{14}{3}$. Multiplying by $3/2$ gives $(1+b)^{3/2} - 1 = 7$, so $(1+b)^{3/2} = 8$. Taking both sides to the power of $2/3$ gives $1+b = 8^{2/3} = (8^{1/3})^2 = 2^2 = 4$. Thus, $b=3$.
14. The derivative of a function $f(x)$ is given by $f'(x) = \sqrt{x^4 - 1}$. Which integral represents the arc length of $y=f(x)$ from $x=1$ to $x=2$?
    Answer: $\int_1^2 x^2 \, dx$ When you encounter an arc length problem, you need to recall the arc length formula for a function $y = f(x)$: $L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx$. This formula measures the actual distance along the curve, accounting for both horizontal and vertical changes. Given that $f'(x) = \sqrt{x^4 - 1}$, you need to substitute this into the arc length formula. First, calculate $[f'(x)]^2 = (\sqrt{x^4 - 1})^2 = x^4 - 1$. Then the arc length integral becomes: $L = \int_1^2 \sqrt{1 + (x^4 - 1)} \, dx = \int_1^2 \sqrt{x^4} \, dx = \int_1^2 x^2 \, dx$ This matches answer choice D. Let's examine why the other answers are incorrect. Choice A gives $\int_1^2 (x^2-1) \, dx$, which completely omits the square root from the arc length formula—this would be incorrect even for a simple area calculation. Choice B shows $\int_1^2 \sqrt{1 + \sqrt{x^4-1}} \, dx$, which incorrectly treats $f'(x)$ as $\sqrt{x^4-1}$ rather than squaring it first; this represents the common mistake of forgetting to square the derivative. Choice C gives $\int_1^2 \sqrt{x^4-2} \, dx$, which appears to result from the error $1 + (x^4 - 1) = x^4 - 2$ instead of the correct $x^4$. Remember: arc length problems always require the formula $\sqrt{1 + [f'(x)]^2}$. Square the given derivative first, then add 1, then take the square root of the entire expression.
15. Let $L$ be the arc length of the curve $y=f(x)$ from $x=a$ to $x=b$. What is the arc length of the curve $y=cf(x/c)$ from $x=ca$ to $x=cb$, where $c > 0$ is a constant?
    Answer: $cL$ Let the new curve be $g(x) = cf(x/c)$. The arc length is given by the integral $\int_{ca}^{cb} \sqrt{1 + [g'(x)]^2} \, dx$. First, we find the derivative of $g(x)$ using the chain rule: $g'(x) = c \cdot f'(x/c) \cdot \frac{1}{c} = f'(x/c)$. Substituting this into the arc length formula gives $\int_{ca}^{cb} \sqrt{1 + [f'(x/c)]^2} \, dx$. We use a u-substitution with $u = x/c$, which means $x = cu$ and $dx = c \, du$. The limits of integration also change: when $x=ca$, $u=a$; when $x=cb$, $u=b$. The integral becomes $\int_a^b \sqrt{1 + [f'(u)]^2} \cdot c \, du = c \int_a^b \sqrt{1 + [f'(u)]^2} \, du$. This is exactly $c$ times the original arc length $L$. Therefore, the new arc length is $cL$.
16. Find the length of one arch of the cycloid given by $x = a(t - \sin t)$ and $y = a(1 - \cos t)$ for $0 \le t \le 2\pi$, where $a>0$.
    Answer: $8a$ First, find the derivatives with respect to $t$: $x'(t) = a(1 - \cos t)$ and $y'(t) = a(\sin t)$. The arc length integrand is $\sqrt{(x')^2 + (y')^2}$. So, $(x')^2 + (y')^2 = a^2(1 - \cos t)^2 + a^2\sin^2 t = a^2(1 - 2\cos t + \cos^2 t + \sin^2 t) = a^2(2 - 2\cos t)$. Using the half-angle identity $1 - \cos t = 2\sin^2(t/2)$, this becomes $a^2(4\sin^2(t/2))$. The integrand is $\sqrt{4a^2\sin^2(t/2)} = 2a|\sin(t/2)|$. For $0 \le t \le 2\pi$, $0 \le t/2 \le \pi$, so $\sin(t/2) \ge 0$. The length is $L = \int_0^{2\pi} 2a\sin(t/2) \, dt = 2a[-2\cos(t/2)]_0^{2\pi} = -4a[\cos(\pi) - \cos(0)] = -4a[-1 - 1] = 8a$.
17. A robot's path is defined by $y = \frac{1}{3}(x^2+2)^{3/2}$. What is the distance the robot travels along this path from $x=0$ to $x=3$?
    Answer: 12 The distance is the arc length. First, find the derivative: $y' = \frac{1}{3} \cdot \frac{3}{2}(x^2+2)^{1/2} \cdot 2x = x\sqrt{x^2+2}$. Next, compute $1+(y')^2$: $1 + (x\sqrt{x^2+2})^2 = 1 + x^2(x^2+2) = 1 + x^4 + 2x^2 = (x^2+1)^2$. The arc length integral is $L = \int_0^3 \sqrt{(x^2+1)^2} \, dx = \int_0^3 (x^2+1) \, dx$. Evaluating the integral gives $[\frac{x^3}{3} + x]_0^3 = (\frac{27}{3} + 3) - 0 = 9 + 3 = 12$.
18. A curve is defined by $x = \sin^3(t)$ and $y = \cos^3(t)$ for $0 \le t \le \pi/2$. What is the length of the curve?
    Answer: 3/2 When you encounter a parametric curve like this, you need the arc length formula for parametric equations: $L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$ First, find the derivatives. For $x = \sin^3(t)$: $\frac{dx}{dt} = 3\sin^2(t)\cos(t)$. For $y = \cos^3(t)$: $\frac{dy}{dt} = -3\cos^2(t)\sin(t)$. Now calculate the expression under the square root: $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 9\sin^4(t)\cos^2(t) + 9\cos^4(t)\sin^2(t) = 9\sin^2(t)\cos^2(t)[\sin^2(t) + \cos^2(t)] = 9\sin^2(t)\cos^2(t)$ Taking the square root: $\sqrt{9\sin^2(t)\cos^2(t)} = 3\sin(t)\cos(t)$ (since both sine and cosine are positive on $[0, \pi/2]$). The arc length becomes: $L = \int_0^{\pi/2} 3\sin(t)\cos(t) \, dt = \frac{3}{2}\int_0^{\pi/2} \sin(2t) \, dt = \frac{3}{2}\left[-\frac{1}{2}\cos(2t)\right]_0^{\pi/2} = \frac{3}{4}[1-(-1)] = \frac{3}{2}$ The answer is D) 3/2. Choice A) 1 likely comes from forgetting the factor of 3 in the derivatives. Choice B) 3/4 results from missing the final factor of 2 when evaluating the definite integral. Choice C) 2 might come from computational errors in the trigonometric integration. Strategy tip: Always factor out constants early in parametric arc length problems, and remember that $\sin(t)\cos(t) = \frac{1}{2}\sin(2t)$ is a key identity for these integrals.
19. Find the arc length of the curve $x = \frac{y^4}{4} + \frac{1}{8y^2}$ from $y=1$ to $y=2$.
    Answer: $123/32$ Since $x$ is given as a function of $y$, we integrate with respect to $y$. The formula is $L = \int_c^d \sqrt{1 + (dx/dy)^2} \, dy$. First, find the derivative: $\frac{dx}{dy} = y^3 - \frac{2}{8y^3} = y^3 - \frac{1}{4y^3}$. Next, compute $1 + (dx/dy)^2$: $1 + (y^3 - \frac{1}{4y^3})^2 = 1 + (y^6 - \frac{1}{2} + \frac{1}{16y^6}) = y^6 + \frac{1}{2} + \frac{1}{16y^6}$. This is a perfect square: $(y^3 + \frac{1}{4y^3})^2$. The integral is $\int_1^2 \sqrt{(y^3 + \frac{1}{4y^3})^2} \, dy = \int_1^2 (y^3 + \frac{1}{4y^3}) \, dy$. Evaluating the integral: $[\frac{y^4}{4} - \frac{1}{8y^2}]_1^2 = (\frac{16}{4} - \frac{1}{8(4)}) - (\frac{1}{4} - \frac{1}{8(1)}) = (4 - \frac{1}{32}) - (\frac{2}{8} - \frac{1}{8}) = 4 - \frac{1}{32} - \frac{1}{8} = 4 - \frac{1}{32} - \frac{4}{32} = 4 - \frac{5}{32} = \frac{128-5}{32} = \frac{123}{32}$.
20. The distance a particle travels along the curve $y=f(x)$ from $x=a$ to $x=b$ is given by $\int_a^b \sqrt{1+\sec^4(x)} \, dx$. What could be $f'(x)$?
    Answer: $\sec^2(x)$ When you encounter an arc length integral, you're looking at the formula $\int_a^b \sqrt{1 + [f'(x)]^2} \, dx$. This formula gives the distance traveled along a curve from point to point. The key is recognizing that whatever appears under the square root as $1 + (\text{something})^2$ tells you that "something" equals $f'(x)$. Looking at the given integral $\int_a^b \sqrt{1+\sec^4(x)} \, dx$, you need to identify what derivative, when squared, gives $\sec^4(x)$. Since $\sec^4(x) = (\sec^2(x))^2$, you have $f'(x) = \sec^2(x)$, which is answer D. Let's check why the other options don't work. For A, if $f'(x) = \sec(x)$, then $[f'(x)]^2 = \sec^2(x)$, giving you $\sqrt{1 + \sec^2(x)}$, not $\sqrt{1 + \sec^4(x)}$. For B, if $f'(x) = \sec(x)\tan(x)$, then $[f'(x)]^2 = \sec^2(x)\tan^2(x)$, which doesn't match either. For C, if $f'(x) = \tan(x)$, then $[f'(x)]^2 = \tan^2(x)$, again not matching. The strategy here is pattern recognition: whenever you see an arc length integral, immediately identify the expression under the square root and work backwards to find what derivative would produce that form when squared. Remember that arc length problems always follow the template $\sqrt{1 + [f'(x)]^2}$.

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