A mode is the most often occurring number, so the answer is 14
It is t divided by r equals 2s
Answer:
Therefore r'(t) =-k sin t i + k cos t j and |r'(t)| = k so T(t) = r'(t)/|r'(t)| = -sin t i + cos t j and T'(t) = -cos t i- sin t j . This gives |T'(t)| = 1, so using this equation, we have κ(t) = |T'(t)|/|r'(t)| = 1/k.
Step-by-step explanation:
We are already given the definition of curvature and the parametrization needed to find the curvature of the circle. In genecral the curvature κ is equal to κ(t)=|T'(t)|/|r'(t)| where r(t) is a parametrization of the curve and T(t) is the normalized tangent vector respect to the parametrization, that is, T(t)=r'(t)/|r'(t)|.
Now, using the derivatives of sines and cosines, and the definition of norm, we obtain that:
r(t) = k cos t i + k sin t j ⇒ r'(t)=-k sin t i + k cos t j ⇒|r'(t)|²=sin²t+cos²t=1
T(t) = r'(t)/|r'(t)|=-sin t i +cos t j ⇒ T'(t)= -cos t i - sin t j ⇒|T'(t)|²=cos²t+sin²t=1
Answer:
15
Step-by-step explanation:
From the frequency table
70 - 79 → 5
80 - 89 → 6
90 - 100 → 4
Students scoring 70 or better = 5 + 6 + 4 = 15
Let's check if the ODE is exact. To do that, we want to show that if
then
. We have
so the equation is indeed exact. We're looking for a solution of the form
. Computing the total differential yields the original ODE,
Integrate both sides of the first PDE with respect to
; then
where
is a function of
alone. Differentiate this with respect to
so that
So the solution to this ODE is
i.e.
