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# 数学代写|微积分代写Calculus Assignment代考|Math222 The arc length function

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## 数学代写|微积分代写Calculus Assignment代考|The arc length function

Consider again the curve of figure 5. Instead of finding the length of the curve from $x=0$ to $x=9$, what if we want to know the length of the curve from $x=0$ to an arbitrary point along the curve-say, at $x=t$ ? The answer is represented in the form of a function, the arc length function.
ARC LENGTH FUNCTION
If $f^{\prime}$ is continuous on an interval and $a$ and $t$ are numbers in that interval, then the arc length function $y=s(t)$ giving the length of the curve $y=f(x)$ from $x=a$ to $x=t$ is
$$s(t)=\int_a^t \sqrt{1+\left(f^{\prime}(x)\right)^2} d x .$$
Example 3 Find the arc length function that gives the length of the curve $y=x^{3 / 2}$ from $x=0$ to $x=t$.

Solution Because the formula for finding the arc length function uses $f^{\prime}(x)$, we first calculate this derivative:
\begin{aligned} y=f(x) &=x^{3 / 2} \ f^{\prime}(x) &=\frac{3}{2} x^{1 / 2} . \end{aligned}
The arc length function is then
\begin{aligned} s(t) &=\int_a^t \sqrt{1+\left(f^{\prime}(x)\right)^2} d x \ &=\int_0^t \sqrt{1+\left(\frac{3}{2} x^{1 / 2}\right)^2} d x \ &=\int_0^t \sqrt{1+\frac{9}{4} x} d x . \end{aligned}

## 数学代写|微积分代写Calculus Assignment代考|Arc length: a strategy that doesn’t wor

At the beginning of this section we developed the formula for arc length using straight line segments between points on the curve, as in figures 2 and 3. The line segments were not all horizontal. By way of contrast, when developing the formula for area under a curve and the formula for volume using the disk method, we approximated the curve with horizontal line segments. Why didn’t we use horizontal segments this time?

Figure 8 illustrates why. If all the line segments are horizontal, then the lengths of the segments always adds up to the total width of the interval, $b-a$. These segments only capture the horizontal width of the curve, completely ignoring any vertical movement in the curve.

Figure 8 A curve $y=f(x)$ (blue) from $x=a$ to $x=b$. The interval $[a, b]$ is partitioned into five subintervals. Horizontal line segments (brown) do not give an approximation to the length of the curve, for the sum of the lengths of horizontal segments is always equal to the width of the interval and does not represent the length of the curve

It is important to remember that although partitioning an interval into infinitely many pieces and forming an omega sum that represents a geometric or physical situation are powerful tools for developing useful formulas, the details of each situation are unique and must be considered carefully. Otherwise, we could end up with a formula that gives us a different quantity than the one we seek.

## 数学代写|微积分代写Calculus Assignment代考|弧长函数

ARC LENGTH FUNCTION
$f^{\prime}$ 在区间上是连续的 $a$ 和 $t$ 在这个区间内的数，是弧长函数吗 $y=s(t)$ 给出曲线的长度 $y=f(x)$ 从 $x=a$ 到 $x=t$
$$s(t)=\int_a^t \sqrt{1+\left(f^{\prime}(x)\right)^2} d x .$$求给出曲线长度的弧长函数 $y=x^{3 / 2}$ 从 $x=0$ 到 $x=t$.

\begin{aligned} y=f(x) &=x^{3 / 2} \ f^{\prime}(x) &=\frac{3}{2} x^{1 / 2} . \end{aligned}

\begin{aligned} s(t) &=\int_a^t \sqrt{1+\left(f^{\prime}(x)\right)^2} d x \ &=\int_0^t \sqrt{1+\left(\frac{3}{2} x^{1 / 2}\right)^2} d x \ &=\int_0^t \sqrt{1+\frac{9}{4} x} d x . \end{aligned}

## MATLAB代写

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