Massively deviating from the given answer on this one, have no idea what's going on. I'm sure that I'm using the correct formulas, but the answer provided is very different from what I come up with.
Find the curvature of the following function:
x = 5cos(t) ; y = 4sin(t) ; t = $pi$/4
Formula for curvature is k = (v(t) x a(t))/$v^3$
r(t) = 5cos(t)i + 4sin(t)j + 0k
v(t) = -5sin(t)i + 4cos(t)j + 0k
a(t) = -5cos(t)i - 4sin(t)j + 0k
v(t) x a(t) = $20sin(t)^2$ + $20cos(t)^2$ k = 20k
|v x a| = 20
That's the numerator, now the denominator:
v = |v| = $\sqrt{(-5sin(t))^2 + (4cos(t))^2}$
= $\sqrt{25sin(t)^2 + 16cos(t)^2}$
With t = $pi/4$:
v = $\sqrt{25sin(pi/4)^2 + 16cos(pi/4)^2}$
= $\sqrt{27.51}$ = $5.25$
So now I have v and |v x a|, so I should be ready to solve:
$20/5.25^3 = 0.138$
Now here's what the book shows:
$40\sqrt{82}/1681 = 0.215$
I can't even begin to fathom where these numbers are coming from, but they are clearly obtained much differently than what I'm doing. Can anyone spot what I'm doing wrong?
The error is in your calculation of the denominator (you wrote down the same error in the numerator as well, but computed it correctly there).
Note that |v| = $\sqrt{(-5sin(t))^2 + (4cos(t))^2} = \sqrt{(-5)^2(sin(t))^2 + (4)^2(cos(t))^2} = \sqrt{25sin^{2}(t) + 16cos^{2}(t)}$ and not $\sqrt{25sin(t)^2 + 16cos(t)^2}$. So by this, $v = \sqrt{20.5} = 4.5277$ approximately. And so, $\dfrac{20}{(4.5277)^3} = 0.215$ approximately, which is the same as the answer in your book.