Answer to Question #154245 in Calculus for mafizur alam

Question #154245

determine the length of x=4(3+y)^2, 1<=y<=4


1
Expert's answer
2021-01-13T15:55:57-0500

x = 4(3 + y)2

    \implies dx /dy = 4 . 2(3 + y) = 24 + 8y

1+ (dx /dy)2 = 1 + (24 + 8y)2

arc length = \int14 1+(dx/dy)2\sqrt{1+ (dx /dy)^2} dy = \int14 1+(24+8y)2\sqrt{1+ (24+8y)^2} dy

substituting u = 24 + 8y

    \implies du = 8dy , when y =1, u =32 and when y = 4 , u = 56


    \implies arc length = \int456 1+(u)2\sqrt{1+ (u)^2} du /8

= [(u1+(u)2\sqrt{1+ (u)^2} + ln(u +1+(u)2\sqrt{1+ (u)^2})/16​​​] 456

=[(561+(56)2\sqrt{1+ (56)^2} + ln(56 +1+(56)2\sqrt{1+ (56)^2})/16​​​]-[(321+(32)2\sqrt{1+ (32)^2} + ln(32 +1+(32)2\sqrt{1+ (32)^2})/16​​​]

=[(563137\sqrt{3137} + ln(56 +3137\sqrt{3137})/16​​​]-[(321025\sqrt{1025} + ln(32 +1025\sqrt{1025})/16​​​]

=132.0349


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