Asked by Alice
                Find d^2y/dx^2 for the curve given by x = 2t + 5 and y= 3t/2
a) 3/(4t+10)
b) (4t+10)/3
c) 1
d) 0
            
        a) 3/(4t+10)
b) (4t+10)/3
c) 1
d) 0
Answers
                    Answered by
            Reiny
            
    x = 2t + 5 
t = (x-5)/2
and y= 3t/2 = (3/2)(x-5)/2 = (3/4)(x - 5)
or
y = (3/4)x - 15/4
so your given parametric equations are simply the straight line
y = (3/4)x - 15/4
dy/dx = 3/4
d(dy/dx) / dx = 0
    
t = (x-5)/2
and y= 3t/2 = (3/2)(x-5)/2 = (3/4)(x - 5)
or
y = (3/4)x - 15/4
so your given parametric equations are simply the straight line
y = (3/4)x - 15/4
dy/dx = 3/4
d(dy/dx) / dx = 0
                    Answered by
            oobleck
            
    Since x and y are both linear functions of t,
dy/dx = (dy/dt) / (dx/dt) will be a constant.
d^2y/dx^2 = 0
    
dy/dx = (dy/dt) / (dx/dt) will be a constant.
d^2y/dx^2 = 0
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