Speed and Velocity: Recall from Calc 1: If f (t) gives the position of an object at time t, then. velocity at time t = f (t) speed at time t = f (t)

Speed and Velocity: Recall from Calc 1: If f (t) gives the position of an object at time t, then velocity at time t = f (t) speed at time t = f (t) Math 36-Multi (Sklensky) In-Class Work January 8, 013 1 / 16

Speed and Velocity: It turns out, the graph of the parametric equations looks like x(t) = t 1 y(t) = t 4 4t. Math 36-Multi (Sklensky) In-Class Work January 8, 013 / 16

Speed and Velocity Example 1, Slope of tangent line when speed is 0: An object s position is given by parametric equations ( t 3 t, t 4 4t 3). Below is a graph of the object s path. When t = 0, object s position is (0, 0) dx dt dy dt t=0 = ( 3t t ) t=0 = 0 t=0 = ( 4t 3 1t ) t=0 = 0 Object comes to brief stop at time t = 0 Slope of tangent line=0: dy dy/dt dx = lim (0,0) t 0 dx/dt = lim t 0 4t 3 1t 3t t Math 36-Multi (Sklensky) In-Class Work January 8, 013 3 / 16

Speed and Velocity Example, Slope of tangent line when speed is 0: An object s position is given by parametric equations ( t 3 3 t, t4 4t ). Below is a graph of the object s path. When t = 1, object s position is ( 1 6, ) dx dt dy dt t=1 = ( t t ) t=1 = 0 t=1 = ( 4t 3 4 ) t=1 = 0 Object comes to brief stop at time t = 1 Slope of tangent line=4: dy dy/dt dx = lim t=1 t 1 dx/dt = lim t 1 4t 3 4 t t = 4 Math 36-Multi (Sklensky) In-Class Work January 8, 013 4 / 16

Recall from reading: Area = b a f (x) dx = b a y dx, if a x b. If working with parametric equations, x = x(t), y = y(t), and if a = x(c) and b = x(d), with c t d, then just replacing y with y(t) and x with x(t) gives us: Be careful: Area = Need a x b and c t d. d c y(t)x (t) dt This is only true if the curve is traversed exactly once from (smaller) t = c to (larger) t = d. Math 36-Multi (Sklensky) In-Class Work January 8, 013 5 / 16

Area of an enclosed region: In Calc 1 or, to find the enclosed area, we would: right endpt left endpt top curve bottom curve dx or top endpt bottom endpt right curve left curve dy Math 36-Multi (Sklensky) In-Class Work January 8, 013 6 / 16

Area of an enclosed region: This loop is traced out by the parametric equations x(t) = t, y(t) = sin(πt), t = 1..1 Math 36-Multi (Sklensky) In-Class Work January 8, 013 7 / 16

Area of an enclosed region: This loop is traced out by the parametric equations x(t) = t, y(t) = sin(πt), t = 1..1 For area, make sure loop is traced out exactly once. On the interval [ 1, 1], x(t) goes from 1 to 0 and back up to 1 again y(t) goes through exactly one period of sine, from 0 down to -1, up to 1, and back to 0 again. Thus on [ 1, 1], the loop is traced out exactly once, clockwise, starting and ending at the point (1, 0). Math 36-Multi (Sklensky) In-Class Work January 8, 013 7 / 16

In Class Work Grab one laptop for your group. =Open Maple. 1. An object s position is given by parametric eqns ( cos(t), sin(3t) ). (a) Use Maple to graph the path of the object from t = π to t = 3π : File - New - Document Mode plot( [ *cos(*t), sin(3*t), t=pi/.. 3*Pi/]) (b) Find the object s position at times t = π and t = π. (c) Find the object s velocity and speed at times t = π, t = π. (d) Describe the object s motion at the given times.. Consider the parametric equations ( 3 cos(t), sin(t) ). (a) Use Maple to graph the curve given by these, on the interval [0, π]. Use similar command to above. If you want to see it to scale, control/right-click on graph, choose Scaling Constrained from pop-up menu. (b) Find the area enclosed by the curve. Return laptop to cart; plug back in. Math 36-Multi (Sklensky) In-Class Work January 8, 013 8 / 16

Solutions to In Class Work 1. An object s position is given by parametric eqns ( cos(t), sin(3t) ). (a) Use Maple to graph the path of the object from t = π to t = 3π : File - New - Document Mode plot( [ *cos(*t), sin(3*t), t=pi/.. 3*Pi/]) Math 36-Multi (Sklensky) In-Class Work January 8, 013 9 / 16

Solutions to In Class Work 1. An object s position is given by parametric eqns ( cos(t), sin(3t) ). (b) Find the object s position at times t = π At time t = π : ( ) π x ( ) π y and t = π. = cos(π) = ( ) 3π = sin = 1 Thus the object is at the point (, 1) at time t = π. At time t = π: x(π) = cos(π) = y(π) = sin(3π) = 0 Thus the object is at the point (, 0) at time t = π Math 36-Multi (Sklensky) In-Class Work January 8, 013 10 / 16

Solutions to In Class Work 1. An object s position is given by parametric eqns ( cos(t), sin(3t) ). (c) Find the object s velocity and speed at times t = π, t = π. At time t = π : Velocity: ( ) x (t) = 4 sin(t) x π = 4 sin(π) = 0 ( ) ( ) y (t) = 3 cos(3t) y π 3π = 3 cos = 0 Thus the velocity is (0, 0) [ ( )] [ ( )] π π Speed = x + y = 0 Math 36-Multi (Sklensky) In-Class Work January 8, 013 11 / 16

Solutions to In Class Work 1. An object s position is given by parametric eqns ( cos(t), sin(3t) ). (c) Find the object s velocity and speed at times t = π, t = π. At time t = π: Velocity: x (t) = 4 sin(t) x (π) = 4 sin(π) = 0 y (t) = 3 cos(3t) y (π) = 3 cos(3π) = 3 The the object s velocity is (0, 3). Speed: Speed = (0) + ( 3) = 3 Math 36-Multi (Sklensky) In-Class Work January 8, 013 1 / 16

Solutions to In Class Work 1. An object s position is given by parametric eqns ( cos(t), sin(3t) ). (d) Describe the object s motion at the given times. At time t = π, the object is briefly at rest. (Before time t = π, the object has been descending along the same path as it is about to ascend. At t = π, it is turning around.) At time t = π, the object is briefly moving straight down, with no horizontal movement. Math 36-Multi (Sklensky) In-Class Work January 8, 013 13 / 16

Solutions to In Class Work. Consider the parametric equations ( 3 cos(t), sin(t) ). (a) Use Maple to graph the curve given by these, on the interval [0, π]. Use similar command to above. If you want to see it to scale, control/right-click on graph, choose Scaling Constrained from pop-up menu. After constraining the scaling, it looks like Math 36-Multi (Sklensky) In-Class Work January 8, 013 14 / 16

Solutions to In Class Work. Consider the parametric equations ( 3 cos(t), sin(t) ). (b) Find the area enclosed by the curve. We need to know if it s being traced out clockwise or counter-clockwise. Either by playing with Maple or by plugging in points, we can see that this is traced out counter-clockwise. Area = - = = 6 t=π t=0 π 0 π 0 y(t)x (t) dt sin(t) 3 sin(t) dt sin(t) sin(t) dt Use the trig identity sin (t) = 1 1 cos(t), or integrate by parts and use a trig identity to evaluate this integral. Math 36-Multi (Sklensky) In-Class Work January 8, 013 15 / 16

Solutions to In Class Work. Consider the parametric equations ( 3 cos(t), sin(t) ). (b) Finding sin (t) dt without using half-angle identity: u = sin(t), dv = sin(t) dt du = cos(t) dt, v = cos(t) sin (t) dt = sin(t) cos(t) + Using that cos (t) = 1 sin (t), cos (t) dt sin (t) dt = sin(t) cos(t) + 1 sin (t) sin (t) dt = t sin(t) cos(t) sin (t) = 1 ( ) t sin(t) cos(t) Math 36-Multi (Sklensky) In-Class Work January 8, 013 16 / 16

Solutions to In Class Work. Consider the parametric equations ( 3 cos(t), sin(t) ). (b) Find the area enclosed by the curve. Area = 6 π 0 sin(t) sin(t) dt [ 1 ( ) ] π = 6 t sin(t) cos(t) 0 ( ) (π ) ( = 3 0 0 0) = 6π Math 36-Multi (Sklensky) In-Class Work January 8, 013 17 / 16