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Mathematics 116
HWK9aSolutions
6.3 p471
Problem 1, §6.3, p471. Use the arc-length formula to find the length of the curve y =
2−3x, −2≤x≤1. Checkyouranswerbynoting that thecurve is a line segment and calculating
its length by the distance formula.
Solution. Since y′ = −3, the length, L, is given by
Z 1 p 2 Z 1 √ √
L= −2 1+(−3) dx= −2 10dx=3 10
To use the distance formula, first note that the two endpoints of the line segment are (−2,8) and
(1,−1). Thus the distance formula gives
p 2 2 √ √
L= (−2−1) +(8+1) = 90=3 10
The two results agree.
3
Problem 8, §6.3, p471. Find the exact length for the curve y = x + 1 , 1 ≤ x ≤ 1.
6 2x 2
Solution. We have
2
′ 2 x 1 2
1+(y ) =1+( − 2)
2 2x
4
=1+x −1+ 1
4
4 2 4x
4
= x +1+ 1
4
4 2 4x
2
x 1 2
= ( + 2)
2 2x
Therefore, the length, L, is given by
Z r Z · ¸
1 2 1 2 3 1
x 1 2 x 1 x 1 31
L= 1 ( + 2) dx= 1 ( + 2)dx= − =
2 2x 2 2x 6 2x 1 48
2 2 2
Problem 18, §6.3, p471. Find the exact length for the curve y = ln(1−y2), 0 ≤ y ≤ 1.
2
′ −2y
Solution. Since x = 1−y2 we have
2 2 2 4 2 2 2
′ 2 4y 4y 1−2y +y +4y (1 +y )
1+(x) =1+ 2 2 = 1+ 2 4 = 2 2 = 2 2
(1−y ) 1−2y +y (1−y ) (1 −y )
Page 1 of 3 A. Sontag October 14, 2001
Math 116 HWK 9a Solns continued
6.3 p471
Therefore, the length, L, is given by
Z 1 s 2 2 Z 1 2 Z 1
L= 2 (1+y ) dy= 2 1+y dy= 2(−1+ 2 )dy
2 2 2 2
0 (1−y ) 0 1−y 0 1−y
(The last step was obtained by using long division.) To integrate the second term, we will need to
use Formula 19 from the text:
Z ¯ ¯
du 1 ¯u+a¯
= ln¯ ¯ +C
2 2 ¯ ¯
a −u 2a u−a
We’ll have y = u,dy = du,a = 1. This gives
Z 1 · ¯ ¯¸1
2 2 2 ¯y +1¯ 2 1
L= (−1+ )dy = −y+ ln¯ ¯ =− +ln3≈0.599
0 1−y2 2 ¯y −1¯ 0 2
Problem 19, §6.3, p471. Find the exact length for the curve y = ln(cosx), 0 ≤ x ≤ π.
4
′ −sinx 2 2
Solution. Since y = cosx = −tanx and since 1+tan x = sec x, the length, L is given by
Z π p Z π Z π
4 4 4
2
L= 0 (secx) dx = 0 |secx|dx = 0 secxdx
To evaluate this last integral, use Formula 14:
Z secudu =ln|secu+tanu|+C
This gives
Z π
4 π √ √
L= secxdx = [ln|secx+tanx|]4 = ln( 2+1)−ln(1+0) = ln( 2+1) ≈ 0.881
0
0
Problem 21, §6.3, p471. Ahawk flying at 15 meters per second at an altitude of 180 meters
accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation
2
y = 180− x
45
Page 2 of 3 A. Sontag October 14, 2001
Math 116 HWK 9a Solns continued
6.3 p471
until its hits the ground, where y is its height above the ground and x is the horizontal distance
traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until
the time it hits the ground.
Solution. When the prey hits the ground, the value of y will be 0, so the prey hits the ground
2
x 2
when 180− 45 = 0 or when x = 45· 180, which gives x = 90. (Note that x has to be positive
here.) So the distance that is traveled by the prey is the same as the arc length for the curve
2
y = 180− x , 0 ≤ x ≤ 90
45
This distance is Z s
90 2
L= 1+ 4x dx
2
0 (45)
Consult the integral tables and plan to use Formula 21:
Z p p 2 p
2 2 u 2 2 a 2 2
a +u du= 2 a +u + 2 ln(u+ a +u )+C
Making the substitution u = 2x,du = 2 dx, and using u = 4 when x = 90 and u = 0 when x = 0,
we find 45 45
Z 90s 2
L= 1+ 4x dx
0 (45)2
45 Z 4p 2
= 2 1+u du
·0 ¸
45 up 1 p 4
2 2
= 2 2 1+u +2ln(u+ 1+u )
0
45 √ 1 √
= 2 [2 17+ 2ln(4+ 17)]
√ 45 √
=45 17+ 4 ln(4+ 17)≈209.1m
Page 3 of 3 A. Sontag October 14, 2001
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