Answer
$\dfrac{4}{25}$
Work Step by Step
Given: $r(t)=3ti+4sintj+4costk$
To calculate the curvature of the curve we will have to use Theorem 10 such as: $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}$
Thus,
$r'(t)=3i+4costj-4sintk$ and $r''(t)=-4sintj-4costk$
and $|r'(t)|=\sqrt {(3)^2+(4cost)^2+(-4sint)^2}$
This implies $|r'(t)|=\sqrt {25}=5$
Now,
$\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{| (3i+4costj-4sintk)\times(-4sintj-4costk|}{|5|^3}$
$=\dfrac{|-16i+12 \cos tj-12\sin tk|}{125}$
$=\dfrac{\sqrt{(-16)^2+(12 \cos t)^2+(12 \sin t)^2}}{125}$
$=\dfrac{20}{125}$
Hence, $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{4}{25}$
