Question

Arc length... http://imgur.com/a/5u8uB

confused on the red portion

Answer 1

Pythagorean Identity states that:
\[||C'(t)||=\sqrt{4^2+3^2}=\sqrt{16+9}=(16+9)^{\frac{ 1 }{ 2 }}=25^{\frac{ 1 }{ 2 }}=5\]

Got it?

I hope that helps!

Answer 2

if you have <4cos4t, 3, -4sin4t> how do you use the pythag id to solve for the magnitude though?

Answer 3

Do you mean how to apply it in 3d system?

Answer 4

I got what I think you are confused about!

Answer 5

For applying the Pythagorean Theorem in the three-dimensional system, it would be the same principle!

Take the square root of the sum of the squared coordinates!
\[\sqrt{(4\cos4t)^2+3^2+(-4\sin4t)^2}=\sqrt{16(\cos4t)^2+9+16(\sin4t)^2}\]
and the most known identity
\[(\cos4t)^2+(\sin4t)^2=1\]

take 16 as a common factor
\[\sqrt{(4\cos4t)^2+3^2+(-4\sin4t)^2}=\sqrt{16[(\cos4t)^2+(\sin4t)^2]+9}\]
\[=\sqrt{16[1]+9}=\sqrt{16+9}=\sqrt{25}=5\]


Done!

Answer 6

@opti

Do you follow?