The graph r(t)=<t-3/2sin(t),t-3/2cos(t),t> . For which values of t is the curvature largest?
I would suggest putting the equation into "desmos" or some other graphing software and obtaining your answer. THEN you can go back to your notes and see how the graph matches up with the equation : )
Is this problem can only be solved by calculator or also by hand and what other graphing software? All I have is a Ti84 calculator.
Recall that for parametric curves,
K = (x'y" - y'x")/(x'^2 + y'^2)^(3/2)
So, plug in your x and y values and find K, then find where dK/dt = 0
Come back if you get stuck, and show what you got.
To find the values of \(t\) for which the curvature is largest, we need to determine the curvature of the graph \(r(t) = \langle t - \frac{3}{2}\sin(t), t - \frac{3}{2}\cos(t), t\rangle\). Let's break down the steps to find the curvature:
Step 1: Calculate the first derivative of \(r(t)\) to find the tangent vector.
The first derivative of \(r(t)\) gives us the tangent vector \(\mathbf{T}\). Since \(r(t) = \langle x(t), y(t), z(t) \rangle\), we can find the tangent vector \(\mathbf{T}\) as follows:
\[
\mathbf{T}(t) = r'(t) = \langle x'(t), y'(t), z'(t) \rangle
\]
Taking the derivative of each component of \(r(t)\) gives:
\[
\begin{align*}
x'(t) &= 1 - \frac{3}{2}\cos(t) \\
y'(t) &= 1 + \frac{3}{2}\sin(t) \\
z'(t) &= 1
\end{align*}
\]
So, the tangent vector \(\mathbf{T}\) is:
\[
\mathbf{T}(t) = \langle 1 - \frac{3}{2}\cos(t), 1 + \frac{3}{2}\sin(t), 1\rangle
\]
Step 2: Calculate the second derivative of \(r(t)\) to find the curvature vector.
The second derivative of \(r(t)\) gives us the curvature vector \(\mathbf{K}\). Taking the derivative of each component of \(\mathbf{T}(t)\) gives:
\[
\begin{align*}
x''(t) &= \frac{3}{2}\sin(t) \\
y''(t) &= \frac{3}{2}\cos(t) \\
z''(t) &= 0
\end{align*}
\]
So, the curvature vector \(\mathbf{K}\) is:
\[
\mathbf{K}(t) = \langle \frac{3}{2}\sin(t), \frac{3}{2}\cos(t), 0\rangle
\]
Step 3: Calculate the magnitude of the curvature vector.
To find the curvature at any point, we need to calculate the magnitude of the curvature vector. The magnitude of the curvature vector \(\mathbf{K}\) is given by:
\[
\|\mathbf{K}(t)\| = \sqrt{\left(\frac{3}{2}\sin(t)\right)^2 + \left(\frac{3}{2}\cos(t)\right)^2 + 0^2} = \frac{3}{2}\sqrt{\sin^2(t) + \cos^2(t)} = \frac{3}{2}
\]
Step 4: Determine the values of \(t\) that make the magnitude of the curvature largest.
Since the magnitude of the curvature \(\|\mathbf{K}(t)\| = \frac{3}{2}\) is a constant, it does not depend on the value of \(t\). Therefore, the curvature is always largest and equal to \(\frac{3}{2}\) for all values of \(t\).
In conclusion, the curvature does not vary with \(t\) and is always largest and equal to \(\frac{3}{2}\).