The curves
r1(t) = 5t, t^2, t^4
&
r2(t) =sin t, sin 4t, 3t
intersect at the origin. Find their angle of intersection, θ, correct to the nearest degree.
funny how much time google can save you
http://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap11/section7/732d51/732_51.html
To find the angle of intersection (θ) between the curves r1(t) and r2(t), we can use the dot product of the tangent vectors at the point of intersection.
The tangent vector of r1(t) is given by r1'(t) and the tangent vector of r2(t) is given by r2'(t).
Let's calculate these tangent vectors:
For r1(t) = 5t, t^2, t^4:
r1'(t) = 5, 2t, 4t^3
For r2(t) = sin t, sin 4t, 3t:
r2'(t) = cos t, 4cos 4t, 3
Since the point of intersection is the origin (0, 0, 0), both curves pass through this point at t = 0.
Now, let's evaluate the tangent vectors at t=0:
r1'(0) = 5, 0, 0
r2'(0) = 1, 4, 3
The dot product of r1'(0) and r2'(0) is given by:
r1'(0) · r2'(0) = (5)(1) + (0)(4) + (0)(3) = 5
To find the angle of intersection (θ), we can use the formula:
cos(θ) = (r1'(0) · r2'(0)) / (|r1'(0)| |r2'(0)|)
|r1'(0)| is the magnitude of the tangent vector r1'(0) which is √(5^2 + 0^2 + 0^2) = √25 = 5
|r2'(0)| is the magnitude of the tangent vector r2'(0) which is √(1^2 + 4^2 + 3^2) = √26
Now we can substitute these values into the formula and solve for cos(θ):
cos(θ) = (r1'(0) · r2'(0)) / (|r1'(0)| |r2'(0)|)
cos(θ) = 5 / (5 * √26)
cos(θ) = 1 / √26
To find θ, we can use the inverse cosine function:
θ = cos^(-1)(1 / √26)
Let's calculate this value:
θ ≈ 1.242 radians
To convert this to degrees, we can multiply by 180/π:
θ ≈ 71 degrees
Therefore, the angle of intersection between the curves r1(t) and r2(t) at the origin is approximately 71 degrees.
To find the angle of intersection between the curves, we can consider the tangent vectors of the curves at the point of intersection (which is the origin in this case). The angle between two vectors is given by the dot product formula:
θ = cos^(-1)((v1 • v2) / (|v1| * |v2|))
where v1 and v2 are the tangent vectors to the curves at the origin.
Let's calculate the tangent vectors:
The tangent vector of r1(t) at the origin can be found by taking the derivative of r1(t) with respect to t and then evaluating it at t = 0:
r1'(t) = (5, 2t, 4t^3)
r1'(0) = (5, 0, 0)
So, the tangent vector of r1(t) at the origin is (5, 0, 0).
Similarly, the tangent vector of r2(t) at the origin can be found by taking the derivative of r2(t) with respect to t and then evaluating it at t = 0:
r2'(t) = (cos t, 4cos 4t, 3)
r2'(0) = (1, 4, 3)
So, the tangent vector of r2(t) at the origin is (1, 4, 3).
Now, let's calculate the angle of intersection (θ) using the dot product formula:
θ = cos^(-1)((v1 • v2) / (|v1| * |v2|))
θ = cos^(-1)((5*1 + 0*4 + 0*3) / (sqrt(5^2 + 0^2 + 0^2) * sqrt(1^2 + 4^2 + 3^2)))
Simplifying the equation further:
θ = cos^(-1)(5 / (sqrt(25) * sqrt(26)))
θ = cos^(-1)(5 / 5 * sqrt(26))
θ = cos^(-1)(1 / sqrt(26))
θ ≈ 63 degrees (rounded to the nearest degree)
Therefore, the angle of intersection between the curves r1(t) and r2(t) at the origin is approximately 63 degrees.