The graph of r(t)=<t- 3sin(t)/2,1- 3cos(t)/2,t> For which values of t is the curvature largest?
I don’t know why my answer is not matching the answer key. Follow up for Oobleck
R’(t)= 1-3cos(t)/2, 3sin(t)/2, 1
R’’(t)=3sin(t)/2,3cos(t)/2,0
This is what I did
Curvature formula K(t)=lf’’(t)l/[1+(f’(t))^2]^(3/2)
r’(t)xr’’(t) ( 1-3cos(t)/2, 3sin(t)/2, 1)x(3sin(t)/2,3cos(t)/2,0)
[(3sin(t)/2*0)- (1* 3cos(t)/2)]i= -3cos(t)/2i
[(1-3cos(t)/2 *0)- ( 3sin(t)/2*1)]j= -3sin(t)/2j
[(1-3cos(t)/2*3cos(t)/2)- (-3sin(t)/2*3sin(t)/2)]k->3cos(t)/2-9cos^2(t)/4- 9sin^2(t)/4k= 3cos(t)/2 -9/4k
lr’(t)xr’’(t)l= √[(-3cos(t)/2)^2+(-3sin(t)/2)^2+(3cos(t)/2 -9/4)^2]
=√(9cos^2(t)/4+9sin^2(t)/4+ 9cos^2(t)/4 -27cos(t)/8 -27cos(t)/8+ 81/16)
=√(9/4cos^2(t)+sin^2(t)+ 9cos^2(t)/4 -54cos(t)/8+ 81/16)
=√(9cos^2(t)/4 -54cos(t)/8+ 81/16+ 9/4)
=√(9cos^2(t)/4 -54cos(t)/8+ 81/16+ 36/16)
lr’(t)xr’’(t)l= √(9cos^2(t)/4 -54cos(t)/8+ 117/16)
[1+(f’(t))^2]^(3/2)= [1+()= [(1-3cos(t)/2)^2+ (3sin(t)/2)^2+( 1)^2]^(3/2)
=[1 -3cos(t)/2 -3cos(t)/2+ 9cos^2(t)/4+ 9sin^2(t)/4) + 1)]^(3/2)
=[-6cos(t)/2+ 9/4cos^2(t)+sin^2(t))+ 1+1)]^(3/2)
=[-6cos(t)/2+ 9/4 + 2)]^(3/2)
=[-6cos(t)/2+ 17/4]^(3/2)
My answer
[√(9cos^2(t)/4 -54cos(t)/8+ 117/16)]/ =[-6cos(t)/2+ 17/4]^(3/2)
Answer key
[√(4cos^2(t)-12cos(t)+13)]/[17-12cos(t)]^(3/2)
typo at these parts
[1+(f’(t))^2]^(3/2)= [1+ (1-3cos(t)/2)^2+ (3sin(t)/2)^2+( 1)^2]^(3/2)
=[1+1 -3cos(t)/2 -3cos(t)/2+ 9cos^2(t)/4+ 9sin^2(t)/4) + 1)]^(3/2)
=[-6cos(t)/2+ 9/4cos^2(t)+sin^2(t))+ 1+1+1)]^(3/2)
=[-6cos(t)/2+ 9/4 + 3)]^(3/2)
=[-6cos(t)/2+ 21/4]^(3/2)
My answer
[√(9cos^2(t)/4 -54cos(t)/8+ 117/16)]/ =[-6cos(t)/2+ 21/4]^(3/2)
Answer key
[√(4cos^2(t)-12cos(t)+13)]/[17-12cos(t)]^(3/2)
The curvature formula you used is for y = f(x).
For general parametric equations, which this is,
k = (x'y" - y'x")/(x'^2 + y'^2)^(3/2) = 6(3-2cost)/(13-12cost)^(3/2)
so, dk/dt = 24sint(3cost-7) / (13-12cost)^(5/2)
dk/dt=0 at all multiples of π
k is a max of 6 at even multiples of π
k is a min of 6/25 at odd multiples of π
To find the values of t for which the curvature is largest, we need to find the maximum of the curvature function K(t).
The curvature formula is K(t) = ||r'(t) x r''(t)|| / ||r'(t)||^3, where r'(t) is the first derivative of r(t), r''(t) is the second derivative of r(t), and ||v|| represents the magnitude of the vector v.
Let's go through the calculations step by step to see where the discrepancy lies.
First, find r'(t):
r'(t) = <1 - 3cos(t)/2, 3sin(t)/2, 1>
Next, find r''(t):
r''(t) = <3sin(t)/2, 3cos(t)/2, 0>
Now, calculate r'(t) x r''(t):
r'(t) x r''(t) = (-3cos(t)/2)(3cos(t)/2) - (3sin(t)/2)(3sin(t)/2) = -9/4*cos^2(t) - 9/4*sin^2(t) = -9/4
The magnitude of r'(t) x r''(t) is ||r'(t) x r''(t)|| = |-9/4| = 9/4.
Next, calculate ||r'(t)||:
||r'(t)|| = sqrt((1 - 3cos(t)/2)^2 + (3sin(t)/2)^2 + 1^2) = sqrt(1 - 3cos(t) + 9cos^2(t)/4 + 9sin^2(t)/4 + 1) = sqrt(9cos^2(t)/4 - 6cos(t) + 17/4)
Finally, calculate K(t):
K(t) = (9/4) / (sqrt(9cos^2(t)/4 - 6cos(t) + 17/4))^3 = (9/4) / (9cos^2(t)/4 - 6cos(t) + 17/4)^(3/2)
Now, let's compare this result with the given answer key:
The given answer key is [√(4cos^2(t) - 12cos(t) + 13)] / [17 - 12cos(t)]^(3/2).
To simplify this expression, notice that √(4cos^2(t) - 12cos(t) + 13) = √[(2cos(t) - 3)^2 + 4] = sqrt(17 - 12cos(t)). Hence, the given answer key can be rewritten as sqrt(17 - 12cos(t)) / [17 - 12cos(t)]^(3/2).
Comparing this with our result, it seems that there is a discrepancy related to the expression inside the square root.
Please recheck your calculations for the vector cross product and the magnitude of r'(t) x r''(t), as well as any simplifications made along the way. Make sure that the vector calculations are executed correctly and that the expressions are simplified accurately.