Calculus w/Parametric Curves
Find dy/dx
x = 2t3+3t
y = 4t-5t2
(4-10t)/(6t2+3)
Find an equation for the line that is parallel to the line
x = 3 - t, y = 6t, z = 7t + 2 and goes through the point P (0, 1, 2).
r(t) = <0,1,2> + <-1,6,7>t
What is the surface corresponding to the equation:
3z = 2x2-12x+15
Cylinder (parabolic)
Find an equation tangent to the curve when t = pi/6
x = sin(t)
y = cos2(t)
y-(3/4) = -(x-1/2) or y = -x+(5/4)
Find parametric equations for the line that goes through the two points A (1,0,-2) and B (4,-2,3)
x = 1 +3t
y = -2t
z = -2+5t
Find the center and radius of the sphere:
x2+y2+z2 = 6x-4y-10z
Center: (3,-2,-5)
Radius: sqrt(38)
Find the area of the region bounded by the parametric curve:
x = t - cos(t) 0<= t <= 2pi
y = sin(t)
pi
Find the equation of the plane that contains the intersecting lines:
x = 4 + t1 x = 4 - 3t2
y = 2t1 y = 3t2
z = 1 - 3t1 z = 1+2t2
13(x-4)+7(y-0)+9(z-1) = 0
(any point on the plane or any scalar multiple of the normal vector works)
See picture of problem:
AE = 23/5
Find d2y/dx2. Simplify completely.
x = ln(sec(t)+tan(t))
y= csc(t) 0<t<pi/2
2cot2(t)csc(t)
Find the equation of the plane that is orthogonal to the plane 3x+2y-z=4 and goes through the points P(1,2,4) and Q(-1,3,2)
3(x-1)-8(y-2)-7(z-4)=0
Let L1 be the line through points A (1,0,0) and B (0,2,1). Let L2 be the line through points C (0,0,2) and D (0,a,0). Find the value of a such that lines L1 and L2 intersect.
a = 4
(the lines intersect at (0,2,1))
Find the EXACT surface area obtained by revolving around the x-axis.
x = 3tet
y = 4tet 0 <= t <= 1
20pie2
Consider the plane that passes through (4,4,2) and contains the line x = 5t, y = 3 + t, z = 4 - t
Find the (x,y,z) point where this plane intersects the y-axis.
(0,11/3,0)
Find an equation of the surface consisting of all points (x,y,z) whose distance to the point (1,1,-1) equals the distance to the plane y = -2. Then identify the surface by name.
(x-1)2 +(z+1)2 = 6y+3
Elliptic paraboloid (it's technically a circular paraboloid)