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A day full of math games & activities. Find one near you.
Find dy/dx and d2y/dx2. x = 3 sin(t), y = 4 cos(t), 0 < t < 2π.
Solution:
Given, x = 3 sin(t), y = 4 cos(t)
We have to find dy/dx and d2y/dx2
dy/dx = (dy/dt)/(dx/dt)
dy/dt = d(4 cos(t))/dt
= -4 sin(t)
dx/dt = d(3 sin(t))/dt
= 3 cos(t)
dy/dx = -4sin(t)/3cos(t)
dy/dx = (-4/3)tan(t)
d2y/dx2 = (d(dy/dx)/dt)/(dx/dt)
d(dy/dx)/dt = d((-4/3)tan(t))/dt
= (-4/3)sec2(t)
d2y/dx2 = (-4/3)sec2(t)/3cos(t)
d2y/dx2 = (-4/9)sec3t
The curve is concave upward when its second derivative is positive.
sec3t is positive for y > 0. So, -sec3t is positive for y < 0.
This implies the curve is concave upward when
y = 4 cos(t) < 0
cos(t) < 0
(π/2) + 2nπ < t < 3π/2 + 2nπ.
Where, n is any integer.
Therefore, dy/dx = (-4/3)tan(t) and d2y/dx2 = (-4/9)sec3t.
Find dy/dx and d2y/dx2. x = 3 sin(t), y = 4 cos(t), 0 < t < 2π.
Summary:
x = 3 sin(t), y = 4 cos(t), 0 < t < 2π. dy/dx = (-4/3)tan(t) and d2y/dx2 = (-4/9)sec3t.
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