Asked by sarah
                find dy/dx and (d^2)y/(d^2)x for x=2sint y=3cost 0<1<2pi
            
            
        Answers
                    Answered by
            Steve
            
    dy/dx = y'/x' = (-3sint)/(2cost) = -3/2 tan t
d^2y/dx^2 = (x'y"-x"y')/x'^3
= ((2cost)(-3cost) - (-2sint)(-3sint))/(8cos^3 t)
= -3/4 sec^3 t
Check:
y = 3/2 √(4-x^2)
y' = -3/2 x/√(4-x^2)
y" = -6/(4-x^2)<sup><sup>3/2</sup></sup>
I'll leave it to you to verify that they are the same.
    
d^2y/dx^2 = (x'y"-x"y')/x'^3
= ((2cost)(-3cost) - (-2sint)(-3sint))/(8cos^3 t)
= -3/4 sec^3 t
Check:
y = 3/2 √(4-x^2)
y' = -3/2 x/√(4-x^2)
y" = -6/(4-x^2)<sup><sup>3/2</sup></sup>
I'll leave it to you to verify that they are the same.
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