Worksheet: Arc Length by Integration

In this worksheet, we will practice using integration to find the length of a curve.

Q1:

Using a trigonometric substitution, determine the arc length of the curve 𝑦=√4βˆ’π‘₯ between π‘₯=0 and π‘₯=π‘˜.

  • A2(π‘˜)arcsin
  • B2ο€½π‘˜2sin
  • C2ο€½π‘˜2arcsin
  • Dsinο€½π‘˜2
  • Earcsinο€½π‘˜2

Q2:

Write the integral required to calculate the length of the sine curve between π‘₯=0 and π‘₯=πœ‹. Do not evaluate it.

  • Aο„Έβˆš1βˆ’π‘₯π‘₯οŽ„οŠ¦οŠ¨sind
  • Bο„Έβˆš1+π‘₯π‘₯οŽ„οŠ¦cosd
  • Cο„Έβˆš1βˆ’π‘₯π‘₯οŽ„οŠ¦οŠ¨cosd
  • D1+π‘₯π‘₯οŽ„οŠ¦οŠ¨sind
  • Eο„Έβˆš1+π‘₯π‘₯οŽ„οŠ¦οŠ¨cosd

Q3:

Calculate the arc length of the curve 𝑦=√4βˆ’π‘₯ between π‘₯=0 and π‘₯=2, giving your answer to 5 decimal places.

Q4:

Find the function 𝑠(π‘₯) which gives the length of the arc 𝑦=√π‘₯ from (0,0) to ο€»π‘₯,√π‘₯ο‡οŠ©.

  • A𝑠(π‘₯)=(9π‘₯+4)βˆ’827
  • B𝑠(π‘₯)=(18π‘₯+8)βˆ’827
  • C𝑠(π‘₯)=4ο€»1+ο‡βˆ’99οŠ―ο—οŠͺ
  • D𝑠(π‘₯)=2ο€»1+ο‡βˆ’33οŠ―ο—οŠͺ
  • E𝑠(π‘₯)=(9π‘₯+4)27

Q5:

Let 𝐹(π‘₯)=ο€»π‘₯,√9βˆ’π‘₯ο‡οŠ¨ on the interval [0,3]. By setting the arc length function 𝑠(𝑑) to be the arc length between 𝐹(0) and 𝐹(𝑑), find the coordinates of the point 𝑃 on this curve such that the arc length from 𝐹(0) to 𝑃 is 1. Give your answer to 4 decimal places.

  • A(0.9816,8.0365)
  • B(0.3272,2.9821)
  • C(1,2.8284)
  • D(0.0175,2.999)
  • E(0.9816,2.8349)

Q6:

The figure shows the curve 𝑦=𝑒+𝑒2ο—οŠ±ο— with marked points 𝐴(1,1.543) and 𝐡(2,3.762).

Use the secant between 𝐴 and 𝐡 to get a lower bound on the length of the curve between those points. Give your answer to 3 decimal places.

Use the three additional points at π‘₯=1.25,1.5,1.75 to get a better approximation of this length to 3 decimal places.

Calculate the length of the curve exactly, giving your answer to 4 decimal places.

Q7:

The figure shows the graph of 𝑦=π‘₯3 with two iterations to estimate the arc length from π‘₯=1 to π‘₯=3 using 2 and then 4 subintervals and adding lengths of the corresponding line segments.

A table of integrals gives the formula ο„Έβˆšπ‘Ž+𝑒𝑒=𝑒2βˆšπ‘Ž+𝑒+π‘Ž2𝑒+βˆšπ‘Ž+𝑒+.dlnC

Use this to compute the exact arc length to 5 decimal places.

Estimate the arc length using the 2 line segments in the first figure, giving your answer to 5 decimal places.

Estimate the arc length using the 4 line segments in the second figure, giving your answer to 5 decimal places.

Using Simpson’s rule with 4 subintervals, the first summand is 16𝑓(1)=√1318. What is the second summand in estimating the arc length? Give your answer to 4 decimal places.

Using Simpson’s rule with 4 subintervals of width 0.5, what is the estimated arc length to 5 decimal places?

Using Simpson’s rule with 8 subintervals of width 0.25, what is the estimated arc length to 5 decimal places?

Q8:

The figure shows a part of the curve 𝐹(π‘₯)=√π‘₯, with marked points 𝐴(0,0), 𝐡, 𝐢, and 𝐷.

Given that the arc length from (0,0) to (π‘₯,𝐹(π‘₯)) is given by 𝑠(π‘₯)=(9π‘₯+4)βˆ’827, the first thing we need for an arc length parameterization is the inverse 𝑠(π‘₯). Determine 𝑓 so that π‘₯=𝑓(𝑠).

  • A𝑓(𝑠)=(18𝑠+8)βˆ’827
  • B𝑓(𝑠)=4ο€»1+ο‡βˆ’99οŠͺ
  • C𝑓(𝑠)=2ο€»1+ο‡βˆ’33οŠͺ
  • D𝑓(𝑠)=π‘ βˆ’49
  • E𝑓(𝑠)=(27𝑠+8)βˆ’49

Hence, give the arc length parameterization π‘₯=𝑓(𝑠), 𝑦=𝑔(𝑠) of the curve.

  • A𝑓(𝑠)=2ο€»1+ο‡βˆ’33οŠͺ, 𝑦=ο„‘ο„£ο„£ο„£ο„£ο„£ο„ βŽ›βŽœβŽœβŽ2ο€»1+ο‡βˆ’33⎞⎟⎟⎠οŠͺ
  • B𝑓(𝑠)=4ο€»1+ο‡βˆ’99οŠͺ, 𝑦=ο„‘ο„£ο„£ο„£ο„£ο„£ο„ βŽ›βŽœβŽœβŽ4ο€»1+ο‡βˆ’99⎞⎟⎟⎠οŠͺ
  • Cπ‘₯=π‘ βˆ’49, 𝑦=ο„‘ο„£ο„£ο„£ο„ οπ‘ βˆ’49
  • D𝑓(𝑠)=(18𝑠+8)βˆ’827, 𝑦=(18𝑠+8)βˆ’827
  • Eπ‘₯=(27𝑠+8)βˆ’49, 𝑦=(27𝑠+8)βˆ’49

The arc length between each of the points 𝐴,𝐡,𝐢, and 𝐷 is one unit. Give the coordinates of 𝐢 to 3 decimal places.

  • A(1.295,1.474)
  • B(1.310,1.500)
  • C(1.295,1.500)
  • D(1.296,1.476)

Q9:

The exact length of the curve 𝑦=𝑒 between π‘₯=1 and π‘₯=2 is 4.785154 to 6 decimal places. Estimate this using the trapezoidal rule with 𝑛=10 for your integral. Give your answer to 6 decimal places.

Q10:

Perform the following.

Find the derivative of βˆšπ‘’+1οŠ¨ο—.

  • A2π‘’βˆšπ‘’+1οŠ¨ο—οŠ¨ο—
  • Bπ‘’βˆšπ‘’+1οŠ¨ο—οŠ¨ο—
  • C𝑒𝑒+1οŠ¨ο—οŠ¨ο—
  • D2βˆšπ‘’+1οŠ¨ο—
  • E1βˆšπ‘’+1οŠ¨ο—

Find the derivative of tanhοŠ±οŠ§οŠ¨ο—ο€»βˆšπ‘’+1.

  • A1βˆšπ‘’+1οŠ¨ο—
  • Bβˆ’1βˆšπ‘’+1οŠ¨ο—
  • Cβˆ’π‘’βˆšπ‘’+1οŠ¨ο—οŠ¨ο—
  • Dπ‘’βˆšπ‘’+1οŠ¨ο—οŠ¨ο—
  • Eβˆ’π‘’βˆšπ‘’+1οŠ¨ο—οŠ¨ο—

Hence, find ο„Έβˆšπ‘’+1οŠ¨ο—.

  • A2βˆšπ‘’+1+ο€»βˆšπ‘’+1+πΆοŠ¨ο—οŠ±οŠ§οŠ¨ο—tanh
  • Bβˆšπ‘’+1+ο€»βˆšπ‘’+1+πΆοŠ¨ο—οŠ±οŠ§οŠ¨ο—tanh
  • Cβˆšπ‘’+1βˆ’ο€»βˆšπ‘’+1+πΆοŠ¨ο—οŠ±οŠ§οŠ¨ο—tanh
  • D2βˆšπ‘’+1βˆ’ο€»βˆšπ‘’+1+πΆοŠ¨ο—οŠ±οŠ§οŠ¨ο—tanh

Use your results above to calculate, to 5 decimal places, the arc length of the curve 𝑦=𝑒 between π‘₯=1 and π‘₯=3.

Q11:

Work out the length of the arc 𝑦=π‘₯+3232π‘₯ between π‘₯=1 and π‘₯=3. Give your answer as a fraction.

  • A319
  • B2249
  • C3.389
  • D6117
  • E6118

Q12:

Calculate the arc length of the curve 𝑦=π‘₯32+1π‘₯οŠͺ between π‘₯=1 and π‘₯=2, giving your answer as a fraction.

  • A932
  • B5132
  • C3932
  • D111160
  • E2332

Q13:

Write, but do not evaluate, an integral for the arc length of the curve 𝑦=ο€Ή2π‘₯+7ln between π‘₯=1 and π‘₯=4.

  • Aο„Έο„ž1+254π‘₯π‘₯οŠͺd
  • Bο„Έο„ž1+25π‘₯π‘₯οŠͺd
  • Cο„Έο„Ÿ1+ο€Ό25π‘₯+7π‘₯π‘₯οŠͺd
  • Dο„Έο„ž1+25π‘₯π‘₯οŠͺd
  • Eο„Έο„ž1βˆ’254π‘₯π‘₯οŠͺd

Q14:

Write, but do not evaluate, an integral for the arc length of the curve π‘₯=3𝑦2tan between 𝑦=0 and 𝑦=πœ‹2.

  • Aο„Έο„ž1+94𝑦2ο‡π‘¦ο‘½οŽ‘οŠ¦οŠ¨secd
  • Bο„Έο„ž1+94𝑦2ο‡π‘¦ο‘½οŽ‘οŠ¦οŠͺsecd
  • Cο„Έο„ž1+94𝑦4ο‡π‘¦ο‘½οŽ‘οŠ¦οŠ¨tand
  • Dο„Έο„ž1+32𝑦4ο‡π‘¦ο‘½οŽ‘οŠ¦οŠ¨secd
  • Eο„Έο„ž1+32ο€»π‘₯2π‘₯ο‘½οŽ‘οŠ¦οŠͺsecd

Q15:

Calculate the arc length of the curve 𝑦=29(3π‘₯βˆ’1) between π‘₯=12 and π‘₯=2, giving your answer as a fraction.

  • A1,111432
  • B9√66
  • C338
  • D7√66
  • E458

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