find the equation of the osculating plane at t=pi/4 on the curve: r(t)=cos(2t)i+sin(2t)j+tk
the osculating plane contains the tangent and the principal normal.
What does your vector analysis text say about this? What do you get?
(r-r0)•TxN = 0
To find the equation of the osculating plane at a given point on a curve, we need to follow these steps:
Step 1: Find the position vector r(t) of the curve.
Step 2: Calculate the velocity vector v(t) by differentiating r(t) with respect to t.
Step 3: Calculate the acceleration vector a(t) by differentiating v(t) with respect to t.
Step 4: Determine the normal vector N(t) by normalizing the cross product of v(t) and a(t).
Step 5: Use the position vector r(t), the velocity vector v(t), and the normal vector N(t) to write the equation of the osculating plane.
Let's work through these steps for the given curve:
Step 1: The position vector of the curve is r(t) = cos(2t)i + sin(2t)j + tk.
Step 2: Differentiating r(t) with respect to t, we get the velocity vector:
v(t) = -2sin(2t)i + 2cos(2t)j + k.
Step 3: Differentiating v(t) with respect to t, we get the acceleration vector:
a(t) = -4cos(2t)i - 4sin(2t)j.
Step 4: Normalizing the cross product of v(t) and a(t), we get the normal vector:
N(t) = (1/4) * (-2sin(2t)(-4sin(2t))i + (-2cos(2t))(-4cos(2t))j + (2cos(2t))(-2sin(2t))k)
= (1/4)(8sin^2(2t)i + 8cos^2(2t)j + 4sin(2t)cos(2t)k)
= 2sin^2(2t)i + 2cos^2(2t)j + sin(2t)cos(2t)k.
Step 5: The equation of the osculating plane at t = pi/4 is given by:
(r - r(pi/4)) · N(pi/4) = 0,
where r(pi/4) is the position vector evaluated at t = pi/4, and N(pi/4) is the normal vector evaluated at t = pi/4.
Substituting the values, we get:
(r - (cos(pi/2)i + sin(pi/2)j + (pi/4)k)) · (2sin^2(pi/2)i + 2cos^2(pi/2)j + sin(pi/2)cos(pi/2)k) = 0
(r - (0i + 1j + (pi/4)k)) · (2i + 2j + 0k) = 0
(r - j - (pi/4)k) · (2i + 2j) = 0.
So, the equation of the osculating plane at t = pi/4 is:
2x + 2y - z - (pi/4) = 0.
To find the equation of the osculating plane at a particular point on a curve, we need to find several vectors that are required to define the plane.
First, we'll find the position vector of the given point on the curve at t = π/4. Let's substitute this value of t into the parametric equation of the curve:
r(π/4) = cos(2(π/4))i + sin(2(π/4))j + (π/4)k
= cos(π/2)i + sin(π/2)j + (π/4)k
= 0i + 1j + (π/4)k
= j + (π/4)k
So, the position vector at t = π/4 is j + (π/4)k.
Next, we need to find the tangent vector at the point. The tangent vector is obtained by differentiating the position vector with respect to t. Let's differentiate the curve equation:
r'(t) = -2sin(2t)i + 2cos(2t)j + k
Now, substitute t = π/4 into the derivative equation:
r'(π/4) = -2sin(2(π/4))i + 2cos(2(π/4))j + k
= -2sin(π/2)i + 2cos(π/2)j + k
= -2i + 2j + k
So, the tangent vector at t = π/4 is -2i + 2j + k.
Finally, to find the normal vector of the osculating plane, we take the cross product of the tangent and binormal vectors:
N = r'(π/4) x r''(π/4)
To find r''(π/4), we differentiate r'(t):
r''(t) = -4cos(2t)i - 4sin(2t)j
Substituting t = π/4:
r''(π/4) = -4cos(2(π/4))i - 4sin(2(π/4))j
= -4cos(π/2)i - 4sin(π/2)j
= 0i - 4j
= -4j
Now, calculate the cross product:
N = (-2i + 2j + k) x (-4j)
= (-2i + 2j + k) x (0i - 4j)
= (-2i + 2j + k) x (-4j)
= (-8k - 8i) - 8j
= -8i - 8j - 8k
Finally, let's find the equation of the osculating plane using the position vector at t = π/4 and the normal vector we just found:
x - (0) = -8i - 8j - 8k dot (j + (π/4)k - (0)i - (0)j - (0)k)
Simplifying the above equation, we get:
x + 8i + 8j + 8k = 0
Therefore, the equation of the osculating plane at t = π/4 is x + 8i + 8j + 8k = 0.