Graph the curve r(t)=(sin(3t),sin(2t),sin(3t)) At how many points on the curve does t appear that the curvature has a local or absolute maximum?
I don't know how to graph this by hand or calculator. Any suggest.
Geogebra.com
Curve(sin(3t),sin(2(t),sin(3t),t,0,2π)
How do I do part b.
Use a CAS to find and graph the curvature function. Does this graph confirm your conclusion from part (a)?
To graph the curve r(t) = (sin(3t), sin(2t), sin(3t), we need to plot points that correspond to different values of t. However, before graphing, let's analyze the curvature of the curve.
The curvature of a curve measures how much the curve deviates from being a straight line at a given point. It measures how fast the tangent vector to the curve is changing. Mathematically, the curvature (κ) of a curve r(t) is given by the formula:
κ = |d(T)/ds|
where T is the unit tangent vector to the curve, s is the arc length parameter, and d(T)/ds is the derivative of T with respect to s.
To find the maximum curvature, we need to find the values of t where the curvature is a local or absolute maximum. To do this, we first differentiate T with respect to s and calculate the norm of this derivative vector.
The unit tangent vector T is given by:
T = r'(t) / |r'(t)|
where r'(t) is the derivative of r(t) with respect to t.
Let's find r'(t):
r'(t) = (3cos(3t), 2cos(2t), 3cos(3t))
Next, we calculate |r'(t)|:
|r'(t)| = sqrt((3cos(3t))^2 + (2cos(2t))^2 + (3cos(3t))^2)
= sqrt(18cos^2(2t) + 18cos^2(3t))
Now, differentiate T with respect to s:
d(T)/ds = d(T)/dt / ds/dt
ds/dt is the magnitude of the velocity vector, which is simply the norm of r'(t). Thus:
ds/dt = |r'(t)| = sqrt(18cos^2(2t) + 18cos^2(3t))
Differentiating T with respect to t gives:
T' = [(3cos(3t))cos(2t) - (2cos(2t))cos(3t),
-(2sin(2t))sin(3t),
(3cos(3t))cos(2t) - (2cos(2t))cos(3t)]
Now, we calculate |d(T)/ds|:
|d(T)/ds| = sqrt((3cos(3t)cos(2t) - 2cos(2t)cos(3t))^2 + (-(2sin(2t))sin(3t))^2 + (3cos(3t)cos(2t) - 2cos(2t)cos(3t))^2)
= sqrt((18(cos^2(3t)cos^2(2t) + cos^2(2t)cos^2(3t)) - 12cos^2(2t)cos^2(3t))^2 + 4sin^2(2t)sin^2(3t))
Now, we have an expression for the curvature κ in terms of t:
κ = |d(T)/ds| = sqrt((18(cos^2(3t)cos^2(2t) + cos^2(2t)cos^2(3t)) - 12cos^2(2t)cos^2(3t))^2 + 4sin^2(2t)sin^2(3t))
To find the values of t where the curvature has a local or absolute maximum, we need to solve the equation κ'(t) = 0, where κ'(t) is the derivative of κ with respect to t. However, given the complexity of the expression for κ, this is not straightforward to solve analytically.
In this case, it would be more practical to use a graphing software or an online graphing tool to plot the curve r(t) and visually identify the points where the curvature is a local or absolute maximum. There are several online graphing tools available (such as Desmos or GeoGebra) that can help you in this regard.
Once you have plotted the curve, look for regions where the curve is highly curved or exhibits sharp turns. These are the points at which the curvature is a local or absolute maximum.