. a. Describe (in words and/or drawing a picture) the curve whose parametric equations are x=5t, y=2sint, ...?

Question

z= 3 cos t

Then find the unit tangent vector at the curve where t= 0

Answer 1

A right conoid ruled surface is formed by moving a straight line intersecting a fixed straight line in such a way that lines are always perpendicular. (See refernce below)

In above case, the x-y plane will form a circle because the components of cos and sin are equal (a=b). This is possible where y = - x+1 type of line revolves. (angle with x axis = 135)

In your case, y=2 sin t ,and z = 3 cost (2 not=3), it make an ellipse (y^2/2 + z^2/2)=1 in y-z plane. For this 2 straight lines or (one straight line of type (x/a + y/b =1) will revolve. I am not where sure as I had done these in 1979-81.

Any way this surface can also be shown as

(3y / 2z) = tan (x /5) - which is called HELICOID (if a=b or 2 = 3)

Parametric equations are:
x = 5t ∂x/∂t = 5
y = 2sint, ∂y∂t = 2cos t ..........(1)
z = 3 cos t ∂z/∂t = - 3 sint
∂u/∂x+y ∂u/∂y
the velocity vector V = i ∂x/∂t + j ∂y/∂t +k ∂z/∂t
= i (5) + j (2 cos t) + k (-3 sin t)

at poit t = 0 (V)t=0 = i (5) + j (2 *1) + k (0) = 5 i + 2 j
mode | V |t=0 = sqrt(29)

the UNIT Tangent parallel to V is

(v)t=o = (V)t=0 / | v | = [1/sqrt(29)] { 5 i + 2 j }

Answer 2

If you let x=0 to see what the projection of this graph is in the yz-plane, you'll see that it is an ellipse (if you do the algebra) with semiminor axis length 4 (along the y axis) and semimajor axis length of 6 (along the z axis). Now, if you project this along the x axis as a linear function of time (which it is), you'll get a elliptical spiral that completes one period every t=2*pi*n (n=+/-0,1,2....). So if you want to see a graph of this function, it completes and elliptical spiral shape first at x=10*pi. To see the complete graph, just repeat that same section over and over into positive and negative infinity along x (since it is a repeating function).

To find the unit tangent vector, r'(t)/abs(r'(t)), you'll need to take the derivative of your vector and find the length of it, then evaluate at t=0

r(t)=<5t, 2sint, 3cost>
r'(t)=<5, 2cost, -3sint>
abs(r'(t))=sqrt(5^2+(2cost)^2+(-3sint)^2)

T(t)=<5, 2cost, -3sint>/(sqrt(5^2+(2cost)^2+(-3sint)^2))
Now evaluate at t=0

T(0)=<5, 2, 0>/(sqrt(5^2+(2)^2))
T(0)=<5, 2, 0>/sqrt(29)

Viola!

Answer 3

you have to eliminate t
y=2sint=2sin(x/5)
hence your curve is y=2sin(x/5)
it would be like a sine wave with period 2(pi)/5 and maximum and minimum values +2,-2 respectively
slope of tangent =dy/dx=2/5cos(x/5) at x=0,slope=2/5

Answer 4

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