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Find the length of the curve
[tex] x = 3y^{4/3} - 3/32y^{2/3} [/tex]
from [tex] -8 \leq y \leq 64 [/tex]
dy/dx = [tex] 4*y^{1/3}-\frac{1}{16y^{1/3}}[/tex] <-- don't need to check this, i used a math tool to check it.
if i plug all of this into a integral, its going to be crazy. so i was wondering if there is a different approach to this problem
[tex] \int_{-8}^{64} \sqrt{1+{(4y^{1/3}-\frac{1}{16y^{1/3}})}^2}[/tex]
[tex] x = 3y^{4/3} - 3/32y^{2/3} [/tex]
from [tex] -8 \leq y \leq 64 [/tex]
dy/dx = [tex] 4*y^{1/3}-\frac{1}{16y^{1/3}}[/tex] <-- don't need to check this, i used a math tool to check it.
if i plug all of this into a integral, its going to be crazy. so i was wondering if there is a different approach to this problem
[tex] \int_{-8}^{64} \sqrt{1+{(4y^{1/3}-\frac{1}{16y^{1/3}})}^2}[/tex]
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