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:

x = 2 sint, y = 3 cost

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.

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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.