Find the Principal unit normal vector to the curve at the specified value of the parameter.
r(t)=ti+t^(2)j+lntk, t=1
r'(t) = i + 2t j + 1/t k
T(t) = r'/|r'| = (i+2tj+1/t k)/√(1+4t^2+1/t^2)
T(1) = (i+2j+k)/√6
the unit normal vector is
N(t) = T'/|T'|
Well, finding the principal unit normal vector to a curve sounds like quite the "normal" task, doesn't it? But fear not, I'm here to make it as "abnormal" as possible!
Now, to find the principal unit normal vector, we need to take the first derivative of the given curve, which is r'(t). Let's get those wheels turning, shall we?
We start by differentiating each component of the curve individually. The derivative of ti with respect to t is simply i, the derivative of t^2j with respect to t is 2tj, and the derivative of lntk with respect to t is (1/t)k.
Now, we can put it all together and calculate the derivative of r(t):
r'(t) = i + 2tj + (1/t)k
Now that we have r'(t), we can find the principal unit normal vector by taking its magnitude.
|r'(t)| = sqrt(1^2 + (2t)^2 + (1/t)^2) = sqrt(1 + 4t^2 + 1/t^2)
Finally, we can determine the principal unit normal vector by dividing r'(t) by its magnitude.
N(t) = (1/sqrt(1 + 4t^2 + 1/t^2))(i + 2tj + (1/t)k)
So, there you have it! The principal unit normal vector to the curve at t = 1 is given by N(1) = (1/sqrt(1 + 4 + 1))(i + 2j + k). I hope that brings a smile to your face, even if it's a "normal" one!
To find the principal unit normal vector to the curve at the specified value of the parameter (t = 1), we need to follow these steps:
Step 1: Calculate the velocity vector, r'(t), of the curve.
Given that r(t) = ti + t^2j + ln(tk), we can differentiate each component of r(t) with respect to t:
r'(t) = i + 2tj + (1/t)k
Step 2: Evaluate r'(t) at t = 1.
Substituting t = 1 into r'(t):
r'(1) = i + 2(1)j + (1/1)k
= i + 2j + k
Step 3: Find the magnitude of r'(1).
The magnitude of a vector is determined by taking the square root of the sum of the squares of its components. In this case, we have:
|r'(1)| = √(1^2 + 2^2 + 1^2)
= √6
Step 4: Calculate the normalized vector by dividing r'(1) by its magnitude.
The normalized vector, denoted as N, is obtained by dividing each component of r'(1) by the magnitude |r'(1)|:
N = (1/√6)i + (2/√6)j + (1/√6)k
= (√6/6)i + (√6/3)j + (√6/6)k
Therefore, the Principal unit normal vector to the curve at t = 1 is (√6/6)i + (√6/3)j + (√6/6)k.
To find the principal unit normal vector to the curve at the specified value of the parameter, we need to follow these steps:
Step 1: Find the velocity vector by differentiating the position vector with respect to t.
Step 2: Find the acceleration vector by differentiating the velocity vector with respect to t.
Step 3: Find the magnitude of the acceleration vector.
Step 4: Divide the acceleration vector by its magnitude to obtain the principal unit normal vector.
Let's go through these steps for the given curve:
Step 1: Find the velocity vector
The position vector r(t) = ti + t^2j + ln(tk).
Differentiating each component with respect to t, we get:
r'(t) = i + 2tj + (1/t)k.
Step 2: Find the acceleration vector
Differentiating the velocity vector r'(t) = i + 2tj + (1/t)k with respect to t, we get:
r''(t) = 2j - (1/t^2)k.
Step 3: Find the magnitude of the acceleration vector
The magnitude of the acceleration vector is given by:
|a(t)| = sqrt((2^2) + (-1/t^2)^2) = sqrt(4 + 1/t^4).
Step 4: Calculate the principal unit normal vector
The principal unit normal vector N(t) is obtained by dividing the acceleration vector by its magnitude:
N(t) = (2j - (1/t^2)k) / sqrt(4 + 1/t^4).
At t=1, we can substitute t=1 into the formula to find the principal unit normal vector at that parameter value.
N(t=1) = (2j - (1/1^2)k) / sqrt(4 + 1/1^4)
= (2j - k) / sqrt(5)
= (0, 2/sqrt(5), -1/sqrt(5)).
Therefore, the principal unit normal vector to the curve at t = 1 is (0, 2/sqrt(5), -1/sqrt(5)).