Find dy/dx and d2y/dx2. For which values of t is the curve concave upward? x = 2sin t, y = 3 cos t, 0 < t < 2.
Solution:
dx/dt = 2 cost
dy/dt = -3 sint
dy/dx = -3 sint/2 cost = (-3/2) tant
d2y/dx2 = \(\frac{\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\mathrm{d} y}{\mathrm{d} x})}{\frac{\mathrm{d}x }{\mathrm{d} t}}\)
d2y/dx2 = \(\frac{\frac{\mathrm{d} }{\mathrm{d} x}(\frac{-3}{2}tant)}{\frac{\mathrm{d}x }{\mathrm{d} t}}\)
d2y/dx2 = (-3/2)sec2t/(2)(cost)
d2y/dx2 = (-3/4)(1/cos3t)
d2y/dx2 = (-3/4)(1/cos3t)
The dy/dx = (-3/2)tant curve is depicted below:
The function dy/dx = (-3/2)tant is concave upwards from π/2 to π and 3π/2 to 2π
The double differential function d2y/dx2 = (-3/4)(1/cos3t) curve is represented below:
The function d2y/dx2 = (-3/4)(1/cos3t) is concave upwards from π/2 to π and 3π/2 to 2π.
Find dy/dx and d2y/dx2. For which values of t is the curve concave upward? x = 2sin t, y = 3 cos t, 0 < t < 2.
Summary:
The dy/dx and d2y/dx2 for the given functions are (-3/2)tant and (-3/4)(1/cos3t) respectively and the values of t for which the curves concave upwards are values which lie within intervals π/2 to π and 3π/2 to 2π.
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