Consider the following parametric equation: x = t^2, y = t^3 - 3t; (a) Find dy/d

Question

Consider the following parametric equation: x = t^2, y = t^3 - 3t; (a) Find dy/dx. Then find the slope and equation of the tangent line to the curve when t = 3. (b) Show that the curve has two tangent lines at points (3, 0) and find their slopes (c) Find the coordinates of the points on the curve where the tangent line is horizontal or vertical (d) Find d^2 y/dx^2 and use it to determine where the curve is concave up and where it is concave down The graphs of the polar curves r_1 = 6 sin 3 theta and r_2 = 3 are shown to the right. the coordinates of the points of intersection curves for 0 lessthanorequalto theta

Explanation / Answer

x=t2 ; y = t3 -3t

dx/dt = 2t; dy/dt = 3t2 - 3 => dy / dx = dy/dt / dx/dt = 3t2 - 3 / 2t

at t=3 , dy/dx = 3* 9 - 3 / 2 * 3 = 24 / 6 = 4

at t=3, x=9; y= 27-9 = 18

Equation of the tangent : y-18 = 4 ( x- 9 ) => 4x-36-y+18 =0 => 4x-y-18=0

b. At (3,0), x=t2 = 3 => t = +/- sqrt (3 )

Since there are two values of t, there are two values of dy/dx and hence two tangents

the two slopes are:

dy/dx |t=-sqrt(3) = 3 ( 3 ) - 3 / 2 * ( -sqrt ( 3) ) = -sqrt( 3)

dy/dx |t=-sqrt(3) = 3 ( 3 ) - 3 / 2 * ( sqrt ( 3) ) = sqrt( 3)

c. When the tangent is horizontal, dy/dx = 0 = > 3t2 - 3 / 2t = 0 => t2 = 1 => t = +/- 1

When the tangent is vertical, dx/dt=0 => 2t= 0 => t=0

d. d2y/dx2 = d/dx ( dy/dx) = d/dt ( dy/dx) / dx/dt = d/dt ( 3t2-3 / 2t ) / 2t = [ 2t ( 6t ) - ( 3t2 - 3) ( 2) ] / ( 2t )2 * 2t

= [12t2 -6t2 + 6 ] / 8t3 = -6t2 +6 / 8t3

for concave up, d2y/dx2 > 0 => 6 - 6t2 > 0 and t3 > 0 => t >0 and t2 < 1 = > 0< t < 1

for concave down, d2y/dx2 < 0 => 6 - 6t2 < 0 and t3 > 0 => t >0 and t2 > 1 = >   t > 1

and 6-6t2 >0 and t3<0 = > t<0 and t2 < 1 => t < 1

Thus for t > 1 and t <1, the curve is concave down,

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