MATH SOLVE

3 months ago

Q:
# Find dy/dx and d2y/dx2. x = 4 sin(t), y = 5 cos(t), 0 < t < 2Ο dy dx = correct: your answer is correct. d2y dx2 = correct: your answer is correct. for which values of t is the curve concave upward? (enter your answer using interval notation.)

Accepted Solution

A:

[tex]x=4\sin t\implies\dfrac{\mathrm dx}{\mathrm dt}=4\cos t[/tex][tex]y=5\cos t\implies\dfrac{\mathrm dy}{\mathrm dt}=-5\sin t[/tex][tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\dfrac{-5\sin t}{4\cos t}=\dfrac{-5\frac44\sin t}{4\cdot\frac55\cos t}=-\dfrac{25}{16}\dfrac xy[/tex]Then it follows that[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{y-x\frac{\mathrm dy}{\mathrm dx}}{y^2}=-\dfrac{25}{16}\dfrac{y^2+x^2}{y^3}[/tex]The curve is concave upward when its second derivative is positive. The numerator [tex]x^2+y^2[/tex] is non-negative for all values of [tex]x,y[/tex], so the denominator determines the sign. [tex]y^3[/tex] is positive for [tex]y>0[/tex], so [tex]-y^3[/tex] is positive for [tex]y<0[/tex]. This means the curve is concave upward when[tex]y=5\cos t<0\implies\cos t<0\implies\dfrac\pi2+2n\pi<t<\dfrac{3\pi}2+2n\pi[/tex]where [tex]n[/tex] is any integer.