Use the formula in Exercise 42 to find the curvature. x=acos(ωt), y=bsin(ωt)
Exercise 42 formula: k = l(x' y'' - y' x'')l / (x'^2 + y'^2)^(3/2)
This is what I did but don't know why my answer not correct.
x'=-aωsin(t), y'=bωcos(ωt),x''=-aω^2cos(ωt), y''=-bω^2sin(ωt)
l[(-aωsin(ωt)*-bω^2sin(ωt)) - (bωcos(ωt)*-aω^2cos(ωt))]l/[(-aωsin(ωt))^2 + (bωcos(ωt))^2]^3/2
l[(abω^3sin^2(ωt)+baω^3cos^2(ωt)]l/[(-a^2ω^2sin^2(ωt) + b^2ω^2cos^2(ωt))]^3/2
labllω^3l/(ω^2)^3/2[-a^2(sin^2(ωt)) + b^2(cos^2(ωt))]^3/2
labllω^3l/ω^3[-a^2(sin^2(ωt)) + b^2(cos^2(ωt))]^3/2
my answer
labl/[-a^2(sin^2(ωt)) + b^2(cos^2(ωt))]^3/2
Why my answer wrong and where did I make mistake at?
you should not have any minus signs in the denominator, since everything is squared.
Your mistake lies in simplifying the expression for curvature. Let's go through it step by step:
Using the given equations for x and y, we can find the first derivatives with respect to t:
x' = -aωsin(ωt)
y' = bωcos(ωt)
Next, we find the second derivatives:
x'' = -aω^2cos(ωt)
y'' = -bω^2sin(ωt)
Next, we substitute the derivatives into the formula for curvature:
k = |x'y'' - y'x''| / (x'^2 + y'^2)^(3/2)
Substituting the derivatives:
k = |-aωsin(ωt)*(-bω^2sin(ωt)) - (bωcos(ωt))*(-aω^2cos(ωt))| / ((-aωsin(ωt))^2 + (bωcos(ωt))^2)^(3/2)
= |abω^3sin^2(ωt) + abω^3cos^2(ωt)| / ((a^2ω^2sin^2(ωt) + b^2ω^2cos^2(ωt)))^(3/2)
Now, we can simplify further using trigonometric identities:
k = |abω^3(sin^2(ωt) + cos^2(ωt))| / ((a^2ω^2sin^2(ωt) + b^2ω^2cos^2(ωt)))^(3/2)
= |abω^3| / (a^2ω^2sin^2(ωt) + b^2ω^2cos^2(ωt))^(3/2)
Therefore, the correct expression for the curvature is:
k = |abω^3| / (a^2ω^2sin^2(ωt) + b^2ω^2cos^2(ωt))^(3/2)
So, your mistake was in omitting the absolute value signs and not simplifying the expression correctly.
To find the curvature using the given formula, let's go through the steps together and identify where the mistake was made.
Step 1: Calculate the derivatives of x and y:
x' = -aωsin(ωt)
y' = bωcos(ωt)
x'' = -aω^2cos(ωt)
y'' = -bω^2sin(ωt)
Step 2: Substitute these values into the formula and simplify:
k = |(x'y'' - y'x'')| / (x'^2 + y'^2)^(3/2)
k = |((-aωsin(ωt))(-bω^2sin(ωt)) - (bωcos(ωt))(-aω^2cos(ωt)))| / ((-aωsin(ωt))^2 + (bωcos(ωt))^2)^(3/2)
k = |(abω^3sin^2(ωt)+baω^3cos^2(ωt))| / (a^2ω^2sin^2(ωt) + b^2ω^2cos^2(ωt))^(3/2)
Step 3: Simplify further:
k = |abω^3sin^2(ωt)+baω^3cos^2(ωt)| / (a^2ω^2sin^2(ωt) + b^2ω^2cos^2(ωt))^(3/2)
Step 4: Factor out ω^2 from both the numerator and the denominator:
k = |ω^2(ab(sin^2(ωt) + cos^2(ωt)))| / (ω^2((a^2sin^2(ωt) + b^2cos^2(ωt)))^(3/2))
Step 5: Simplify using the trigonometric identity sin^2(ωt) + cos^2(ωt) = 1:
k = |ω^2(ab)| / (ω^2((a^2sin^2(ωt) + b^2cos^2(ωt)))^(3/2))
k = |ab| / ((a^2sin^2(ωt) + b^2cos^2(ωt))^(3/2))
The correct answer for the curvature is |ab| / ((a^2sin^2(ωt) + b^2cos^2(ωt))^(3/2)). It seems that you missed the ω^2 term in the denominator and didn't eliminate the square root entirely.
I hope this clears up any confusion and helps you understand the correct solution. Let me know if you have any further questions!