📝 Arc length of a curve calculus (30 MCQs)

📖 Explanation: The student mistakenly simplified $\sqrt{1+(2x)^2}$ as $1+2x$. The correct integrand is $\sqrt{1+4x^2}$. Squaring a sum is not the same as the sum of squares. This is a common algebraic error when applying the arc length formula.

📖 Explanation: The arc length formula for $y=f(x)$ is \int_a^b \sqrt{1+[f'(x)]^2}\,dx. For $y=\sin x$, f'(x)=\cos x. Squaring it gives $\cos^2 x$, so the integrand is $\sqrt{1+\cos^2 x}$. Easy of the standard formula is required.

📖 Explanation: When a curve is expressed as $x=g(y)$, the formula for arc length from $y=c$ to $y=d$ is \int_c^d \sqrt{1+[g'(y)]^2}\,dy. The derivative must be with respect to $y$, and the limits must be in terms of $y$. Easy is needed here.

📖 Explanation: The derivation of the arc length formula uses the Mean Value Theorem to express the change in $y$ over a subinterval as f'(x_k^*)\Delta x_k. This is crucial for transforming the sum of segment lengths into a Riemann sum that can be evaluated as an integral. It's not about continuity or existence of the limit in a general sense.

📖 Explanation: The arc length formula is derived assuming $y=f(x)$ is a smooth curve, which requires f' to be continuous on $[a,b]$. This ensures the Mean Value Theorem can be applied and the resulting integral exists. Continuity of $f$ alone is not sufficient for the derivation to work.

📖 Explanation: The derivative of $y=x^{3/2}$ is $\frac{3}{2}x^{1/2}$. Squaring this gives $\frac{9}{4}x$. The correct integrand is $\sqrt{1+\frac{9}{4}x}$. Options B, C, and D have errors in the exponent or the derivative. Medium of the formula requires careful differentiation and algebra.

📖 Explanation: For $y=\cosh x$, y'=\sinh x, so the arc length integrand is $\sqrt{1+\sinh^2 x}$. Using the identity $\cosh^2 x - \sinh^2 x=1$, this simplifies to $\sqrt{\cosh^2 x}=\cosh x$. This is the core conceptual step that makes the integral easy to evaluate. The derivative property is part of it, but the identity is key to simplification.

📖 Explanation: The curve $y=x^2$ from $x=1$ to $2$ is the same as the curve $x=\sqrt{y}$ from $y=1$ to $4$. The arc length is a geometric property of the curve and is independent of the parameterization. Both integrals, after appropriate substitution (e.g., $u=x^2$), should evaluate to the same numerical value, demonstrating the equivalence of the two formulations.

📖 Explanation: This requires Easy that arc length depends on the magnitude of the derivative. For $y=e^x-1$, y'=e^x, which is large and grows, leading to a larger integrand. While each integral would need to be evaluated, understanding that a larger derivative over the interval generally leads to a larger arc length is key. Option D has the most significant variation in height, suggesting a longer curve.

📖 Explanation: Arc length considers the total distance traveled along the curve. The line segment connecting two points has both a horizontal change ($\Delta x_k$) and a vertical change ($\Delta y_k$). Ignoring $\Delta x_k$ would only measure the vertical component, which would be incorrect unless the curve is vertical. The formula uses the Pythagorean theorem to find the true length of the segment.

📖 Explanation: For $y=2f(x)$, the derivative is 2f'(x). The arc length is \int_a^b \sqrt{1+[2f'(x)]^2}\,dx = \int_a^b \sqrt{1+4[f'(x)]^2}\,dx. It is not simply $2L$ because the vertical scaling also affects the horizontal component of the curve's slope, changing the length in a non-linear way.

📖 Explanation: This problem requires multiple conceptual steps. First, find y'=\tan x. Then, the integrand becomes $\sqrt{1+\tan^2 x}$. Since $\sec x$ is positive on the interval, $\sqrt{\sec^2 x}=\sec x$. Each step (differentiation, identity, and considering the sign) is crucial and interdependent, making this a Hard problem that tests comprehensive understanding.

📖 Explanation: Adding a constant to a function translates it vertically. The derivative, f'(x), remains unchanged. Since the arc length formula depends only on the derivative, the arc length remains the same. A vertical shift does not change the shape or length of the curve. This tests a deep understanding of the formula's dependence on the derivative.

📖 Explanation: For $y=x^{2/3}$, the derivative is y'=\frac{2}{3}x^{-1/3}, which is undefined at $x=0$. The arc length formula requires f' to be continuous. The cusp at $x=0$ makes the curve not smooth, and the integral would be improper. A common error is to apply the formula without checking the smoothness condition.

📖 Explanation: For parametric curves, the arc length formula is $\int_a^b \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\,dt$. Here, $\frac{dx}{dt}=2t$ and $\frac{dy}{dt}=3t^2$. Squaring and adding gives $4t^2+9t^4$. This is a direct Medium of the parametric arc length formula, requiring careful differentiation and algebra.

📖 Explanation: Increasing the number of line segments (subintervals) generally provides a better approximation to the curve, as the polygonal path more closely follows the curve. However, this is only true if the segments are chosen appropriately. In theory, as the number of segments approaches infinity, the approximation approaches the exact arc length. This tests understanding of the limiting process.

📖 Explanation: The arc length integral is the limit of the sum of lengths of the line segments that approximate the curve. It is not an area, nor is it simply the total change in $y$. This definition is fundamental to understanding the geometric meaning of the integral in this context. It formalizes the intuitive idea of measuring a curve by straightening it out.

📖 Explanation: The formula does not require the function to be one-to-one. It only requires f' to be continuous (smooth). A common misconception is that invertibility is needed, but the formula works for any smooth curve, even those that are not one-to-one, as it sums the lengths of the segments regardless of whether the curve goes 'backwards' horizontally.

📖 Explanation: A vertical shift of the curve does not change its derivative. Since the arc length depends only on f'(x), the length remains $\sinh a$. The student incorrectly assumes the shift adds to the length. This is a misconception about the effect of translations on the curve's length, which only changes if the shift is in the direction of the independent variable.

📖 Explanation: The integral $\int \sqrt{1+\frac{1}{x^2}} dx = \int \frac{\sqrt{x^2+1}}{x} dx$ can be simplified with $u=\sqrt{x^2+1}$. This substitution is non-trivial and represents a higher-level Medium of integration techniques combined with arc length. It tests the ability to recognize the structure of the integrand and apply a suitable substitution.

📖 Explanation: This problem tests the effect of a horizontal scaling on arc length. If $y=f(2x)$, then y'=2f'(2x). The arc length from 0 to 1 is \int_0^1 \sqrt{1+4[f'(2x)]^2}\,dx. This is not simply a factor of $L$, as the scaling changes the slope and the horizontal component of length. This requires a deep understanding of the arc length formula and transformations.

📖 Explanation: If the curve is symmetric about the line $x=2$, the shape of the curve from $x=2$ to $x=4$ is a mirror image of the curve from $x=0$ to $x=2$. The length of a mirrored curve segment is identical to the original segment's length. This is a purely geometric reasoning problem based on graph symmetry and does not require integration.

📖 Explanation: $y=\sqrt{9-x^2}$ is the upper half of a circle of radius 3 centered at the origin. The arc from $x=0$ to $x=3$ is a quarter of the circle. Its length is $\frac{1}{4}(2\pi(3)) = \frac{3\pi}{2}$. This is much more efficient than setting up and evaluating the integral. This tests the ability to recognize geometric shapes from their equations.

📖 Explanation: The second curve is also a quarter circle, but the parameter $t$ runs from 0 to $\pi/4$, while the angle $\theta=2t$ runs from 0 to $\pi/2$. The arc length of a quarter circle of radius 1 is always $\pi/2$, regardless of the parametrization's speed, as long as the trace is exactly the quarter circle once. This tests the understanding that arc length is a geometric property independent of parametrization.

📖 Explanation: With $u=e^x$, $du=e^x dx \implies dx = du/u$. The limits become $u=1$ to $u=e$. The integrand becomes $\sqrt{1+u^2}\cdot \frac{du}{u}$. This is a correct Medium of substitution, requiring careful handling of the $dx$ term and the limits. It tests a multi-step Medium of calculus techniques.

📖 Explanation: For $y=\frac{1}{x}$, y'=-\frac{1}{x^2}. Squaring gives $\frac{1}{x^4}$. The integrand is $\sqrt{1+\frac{1}{x^4}}$. This requires recognizing the derivative of $1/x$ and working backward from the integrand. It tests the ability to reverse-engineer a curve from its arc length integral.

📖 Explanation: While the arc length formula gives an exact definition, many resulting integrals (e.g., $\int_0^1 \sqrt{1+x^3}\,dx$) have no elementary antiderivative. Numerical integration is the practical way to compute these lengths. This highlights a key limitation of the symbolic integration techniques learned in calculus and the importance of numerical methods.

📖 Explanation: The inverse function $x=f^{-1}(y)$ is the exact same curve as $y=f(x)$, just viewed with the axes swapped. The arc length is a geometric property of the curve and does not depend on the coordinate system. Therefore, the length is unchanged. This is a high-level conceptual question linking arc length to the geometric idea of a curve, independent of its parameterization.

📖 Explanation: Since ds = \sqrt{1+[f'(x)]^2} dx, we have \frac{dx}{ds} = \frac{1}{\sqrt{1+[f'(x)]^2}}. Then \frac{dy}{ds} = \frac{dy}{dx}\frac{dx}{ds} = \frac{f'(x)}{\sqrt{1+[f'(x)]^2}}. This is a sophisticated problem that requires understanding the relationship between the differential of arc length and the standard derivative, going beyond basic formula Medium.

📖 Explanation: A vertical tangent means the slope $dy/dx$ is infinite. The formula $y=f(x)$ requires a finite derivative. The curve can often be re-parameterized, for example, as $x=g(y)$, to handle this. A student might incorrectly assume the length is infinite or that the function is discontinuous. The problem is with the representation, not the curve itself.