Find the length of the arc of the semicubical parabola y2 = x between the points...

 

 

 

Transcribed Image Text:

Find the length of the arc of the semicubical parabola y2 = x³ between the points (1, 1) and (16, 64). (See the figure.) SOLUTION For the top half of the curve we have y=x³/2 dy 3 and so the arc length formula gives V1+ 64 16 x dx. If we substitute u- then du - dx. When x = 1, L = 16 16 - [₁0 √ ₁ + (OX) ² αx = [² dx ; when x = 16, u= Vu du 137 13/4 Therefore, As a check on our answer to this example, notice from the figure that the arc length ought to be slightly larger than the distance from (1, 1) to (16, 64), which is √4194-64.761099 According to our calculation in the example, we have (373/2 (13/4)3/2) - 64.949094 Sure enough, this is a bit greater than the length of the line segment. Find the length of the arc of the semicubical parabola y2 = x³ between the points (1, 1) and (16, 64). (See the figure.) SOLUTION For the top half of the curve we have y=x³/2 dy 3 and so the arc length formula gives V1+ 64 16 x dx. If we substitute u- then du - dx. When x = 1, L = 16 16 - [₁0 √ ₁ + (OX) ² αx = [² dx ; when x = 16, u= Vu du 137 13/4 Therefore, As a check on our answer to this example, notice from the figure that the arc length ought to be slightly larger than the distance from (1, 1) to (16, 64), which is √4194-64.761099 According to our calculation in the example, we have (373/2 (13/4)3/2) - 64.949094 Sure enough, this is a bit greater than the length of the line segment.

Answer rating: 100% (QA)

Given that are length let Lower limit upper limit 37 L ... View the full answer