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Explain what is wrong with the statement. The arc length of the curve \(y=\sin x\) from \(x=0\) to \(x=\pi / 4\) is \(\int_{0}^{\pi / 4} \sqrt{1+\sin ^{2} x} d x\).

Short Answer

Expert verified
The formula uses \( \sin^2 x \) instead of the correct \( \cos^2 x \).

Step by step solution

01

Understanding Arc Length Formula

The arc length of a curve given by a function \( y = f(x) \) from a point \( a \) to \( b \) is calculated using the formula: \( \int_a^b \sqrt{1 + (f'(x))^2} \, dx \). This formula uses the derivative of the function.
02

Find the Derivative of the Function

Determine the derivative \( f'(x) \) of the function \( y = \sin x \). The derivative is \( f'(x) = \cos x \).
03

Substitute Derivative into Arc Length Formula

Substitute \( f'(x) = \cos x \) into the arc length formula, resulting in: \( \int_0^{\pi/4} \sqrt{1 + \cos^2 x} \, dx \).
04

Identify the Error in the Given Formula

The given statement uses \( \int_0^{\pi/4} \sqrt{1 + \sin^2 x} \, dx \), but it should be \( \int_0^{\pi/4} \sqrt{1 + \cos^2 x} \, dx \) based on the derivative.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arc Length
Arc length is a concept in geometry that helps calculate the distance along a curved line. This is especially useful when dealing with graphs of functions which are not linear. To find the arc length of a continuous curve described by a function, we use the formula:\[L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx\]Here, \( L \) represents the arc length, while \( f'(x) \) indicates the derivative of the function \( f(x) \) between the limits \( a \) and \( b \). The term \( \sqrt{1 + (f'(x))^2} \) takes into account the inclination of the curve at different points. This formula effectively measures the direct distance along the curve itself, not just horizontally. Calculating arc length allows you to understand how a function behaves over an interval, showing the true path length.
Integral Calculus
Integral calculus is a central theme in mathematics that deals with the accumulation of quantities and finding areas under curves. It involves the process of integration, which is essentially the reverse operation of taking a derivative. One of the primary applications of integral calculus is determining areas, volumes, and sometimes lengths. The definite integral \( \int_a^b f(x) \, dx \) computes the net area under the curve \( f(x) \) from \( x = a \) to \( x = b \). In arc length problems, integration helps in summing up the tiny segments of lengths along a curve. Integral calculus becomes particularly powerful when combined with the concept of limits. It allows for precise calculations of quantities that would be too complex or impossible to measure directly.
Derivative Function
The derivative of a function represents the rate at which a function's value changes with respect to a change in one of its variables. For a function \( y = f(x) \), the derivative \( f'(x) \), conveys the slope of the tangent line to the graph of the function at any point \( x \). Understanding derivatives is crucial when calculating arc length because of the need to know how the function's direction changes.Calculating the derivative involves finding \( f'(x) \) using rules of differentiation, such as the power rule, product rule, or chain rule. For example, the derivative of \( y = \sin x \) is \( f'(x) = \cos x \). This derivative reflects how the sine function rises and falls within one cycle of its curve.The derivative helps in determining the curvature of the graph and in calculating the exact arc length of a specific section of a curve.

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Most popular questions from this chapter

A hemisphere of radius \(a\) has its base on the \(x y\) -plane, centered at the origin; the \(z\) -axis is vertically upward. Using the given slices, (a) Write an expression for the volume of a slice. (b) Write an integral giving the volume of the hemisphere. (c) Calculate the integral. Vertical slices perpendicular to the \(x\) -axis.

Consider the region bounded by \(y=e^{x}\) the \(x\) -axis, and the lines \(x=0\) and \(x=1 .\) Find the volume of the solid. The solid obtained by rotating the region about the horizontal line \(y=7\).

Climbing a ladder at a constant rate, a painter sprays \(1 / 4 \mathrm{kg}\) of paint per meter onto a pole from the ground up to a height of 8 meters. Her sprayer starts on the ground with \(3 \mathrm{kg}\) of paint and weighs \(2 \mathrm{kg}\) when empty. (a) Find a formula for the mass of the sprayer with paint as a function of \(h,\) the height of the sprayer above the ground. (b) Approximate the work done by the painter in lifting the sprayer from height \(h\) to \(h+\Delta h.\) (c) Find total work done lifting the sprayer for one coat of paint.

Old houses may contain asbestos, now known to be dangerous; removal requires using a special vacuum. A contractor climbs a ladder and sucks up asbestos at a constant rate from a \(10 \mathrm{m}\) tall pipe covered by \(0.2 \mathrm{kg} / \mathrm{m}\) using a vacuum weighing \(14 \mathrm{kg}\) with a \(1.2 \mathrm{kg}\) capacity. (a) Let \(h\) be the height of the vacuum from the ground. If the vacuum is empty at \(h=0,\) find a formula for the mass of the vacuum and the asbestos inside as a function of \(h.\) (b) Approximate the work done by the contractor in lifting the vacuum from height \(h\) to \(h+\Delta h\) (c) At what height does the vacuum fill up? (d) Find total work done lifting the vacuum from height \(h=0\) until the vacuum fills. (e) Assuming again an empty tank at \(h=0,\) find the work done lifting the vacuum when removing the remaining asbestos.

Explain what is wrong with the statement. A cylindrical tank is 10 meters deep. It takes twice as much work to pump all the oil out through the top of the tank when the tank is full as when the tank is half full.

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