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Comparison of Approximate and Exact Length of Curve

Created by @nathanedwards
 at November 3rd 2023, 6:53:14 am.
AP_Calculus_AB-Questions
#ap-calculus-ab
#curve-length
#approximation

AP Calculus AB Exam Question

The length of a curve y=f(x)y=f(x) between x=ax=a and x=bx=b can be approximated by dividing the interval [a,b][a,b] into nn equal subintervals and summing the lengths of the line segments connecting the points (xi,yi)(x_i, y_i) and (xi+1,yi+1)(x_{i+1}, y_{i+1}), where xi=a+in(b−a)x_i = a + \frac{i}{n}(b-a) and yi=f(xi)y_i = f(x_i).

Consider the curve y=2x2−3xy = 2x^2 - 3x between x=0x=0 and x=3x=3.

a) Divide the interval [0,3][0,3] into four equal subintervals. Compute the length of the curve using the given approximating method.

b) Determine an exact expression for the length of the curve y=2x2−3xy = 2x^2 - 3x between x=0x=0 and x=3x=3.

c) Compare the exact length in part b) with the approximate length in part a). Discuss the accuracy of the approximation.

Note: You may leave your answer in terms of integers, fractions, and simplified radicals.

Answer:

a) To divide the interval [0,3][0,3] into four equal subintervals, we need to find the values of xix_i for i=0,1,...,4i=0,1,...,4 using the formula:

xi=a+in(b−a)x_i = a + \frac{i}{n}(b-a)

For this question, a=0a=0, b=3b=3, and n=4n=4. Plugging them into the formula, we get:

x0=0+04(3−0)=0x_0 = 0 + \frac{0}{4}(3-0) = 0
x1=0+14(3−0)=34x_1 = 0 + \frac{1}{4}(3-0) = \frac{3}{4}
x2=0+24(3−0)=64=32x_2 = 0 + \frac{2}{4}(3-0) = \frac{6}{4} = \frac{3}{2}
x3=0+34(3−0)=94x_3 = 0 + \frac{3}{4}(3-0) = \frac{9}{4}
x4=0+44(3−0)=3x_4 = 0 + \frac{4}{4}(3-0) = 3

Next, we need to compute the corresponding yy-values yi=f(xi)y_i = f(x_i) for each xix_i:

y0=f(x0)=f(0)=2(0)2−3(0)=0y_0 = f(x_0) = f(0) = 2(0)^2 - 3(0) = 0
y1=f(x1)=f(34)=2(34)2−3(34)=−98y_1 = f(x_1) = f\left(\frac{3}{4}\right) = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) = -\frac{9}{8}
y2=f(x2)=f(32)=2(32)2−3(32)=−92y_2 = f(x_2) = f\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) = -\frac{9}{2}
y3=f(x3)=f(94)=2(94)2−3(94)=−98y_3 = f(x_3) = f\left(\frac{9}{4}\right) = 2\left(\frac{9}{4}\right)^2 - 3\left(\frac{9}{4}\right) = -\frac{9}{8}
y4=f(x4)=f(3)=2(3)2−3(3)=9y_4 = f(x_4) = f(3) = 2(3)^2 - 3(3) = 9

Now, we can compute the length of the curve by summing the lengths of the line segments connecting adjacent points using the distance formula:

L≈∑i=0n−1(xi+1−xi)2+(yi+1−yi)2L \approx \sum_{i=0}^{n-1} \sqrt{(x_{i+1} - x_i)^2 + (y_{i+1} - y_i)^2}

Substituting the values we found, we have:

L≈(34−0)2+(−98−0)2+(32−34)2+(−92−(−98))2+(94−32)2+(−98−(−92))2+(3−94)2+(9−(−98))2L \approx \sqrt{\left(\frac{3}{4} - 0\right)^2 + \left(-\frac{9}{8} - 0\right)^2} + \sqrt{\left(\frac{3}{2} - \frac{3}{4}\right)^2 + \left(-\frac{9}{2} - (-\frac{9}{8})\right)^2} + \sqrt{\left(\frac{9}{4} - \frac{3}{2}\right)^2 + \left(-\frac{9}{8} - (-\frac{9}{2})\right)^2} + \sqrt{\left(3 - \frac{9}{4}\right)^2 + (9 - (-\frac{9}{8}))^2}

Simplifying each square root term, we get:

L≈916+8164+3616+78+916+22564+14416+15364L \approx \sqrt{\frac{9}{16} + \frac{81}{64}} + \sqrt{\frac{36}{16} + \frac{7}{8}} + \sqrt{\frac{9}{16} + \frac{225}{64}} + \sqrt{\frac{144}{16} + \frac{153}{64}}
L≈9064+558+23464+29764L \approx \sqrt{\frac{90}{64}} + \sqrt{\frac{55}{8}} + \sqrt{\frac{234}{64}} + \sqrt{\frac{297}{64}}
L≈3104+552+3394+3334L \approx \frac{3\sqrt{10}}{4} + \frac{\sqrt{55}}{2} + \frac{3\sqrt{39}}{4} + \frac{3\sqrt{33}}{4}

b) To determine the exact expression for the length of the curve y=2x2−3xy = 2x^2 - 3x between x=0x=0 and x=3x=3, we need to find the integral of 1+(dydx)2\sqrt{1 + \left(\frac{dy}{dx}\right)^2} with respect to xx over the interval [0,3][0,3].

First, let's find dydx\frac{dy}{dx}:

dydx=ddx(2x2−3x)=4x−3\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 3x) = 4x - 3

Then, the exact length is given by the integral:

L=∫031+(dydx)2 dxL = \int_{0}^{3} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
L=∫031+(4x−3)2 dxL = \int_{0}^{3} \sqrt{1 + (4x - 3)^2} \, dx

To evaluate this integral, we can use a numerical method or a calculator, which gives us:

L≈5.190L \approx 5.190

c) Comparing the exact length in part b) with the approximate length in part a), we have:

Exact Length = 5.190 (approx.)

Approximate Length = 3104+552+3394+3334\frac{3\sqrt{10}}{4} + \frac{\sqrt{55}}{2} + \frac{3\sqrt{39}}{4} + \frac{3\sqrt{33}}{4} (approx.)

While the exact length is given to three decimal places, the approximate length is expressed as a sum of radicals. Therefore, the approximation is less accurate compared to the exact length. However, the approximation method still provides a reasonable estimate for the length of the curve.

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