# The curve vec{r}(t) = {t, t, t^2} has one point with maximal curvature. Find this point and...

## Question:

The curve {eq}\vec{r}(t) = \left \langle {t, t, t^2} \right \rangle {/eq} has one point with maximal curvature. Find this point and its curvature.

## Maximal Curvature:

The function for curvature, {eq}\kappa (x) = \frac{|f''(x)|}{(1 + (f'(x))^2)^{\frac{3}{2}}} {/eq}. Once we have found the first and second derivatives of our original function, {eq}f(x) {/eq}, we substitute them into the curvature function. Then, we find the critical points of {eq}\kappa (x) {/eq} by finding the zeros of the first derivative. To classify the critical points, we use the first or second derivative test. Then, we find the corresponding curvature value of the maximum and minimum.

## Answer and Explanation:

To find the curvature function, we must first find the first and second derivatives of our original function, {eq}f(t) = t^2 {/eq}.

$$\begin{align*} f(t) &= t^2 \\ f'(t) &= 2t \\ f''(t) &= 2 \end{align*} $$

Using the first and second derivative equations, we generate the curvature function.

$$\begin{align*} \kappa (t) &= \dfrac{|f''(t)|}{(1 + (f'(t))^2)^{\frac{3}{2}}} \\ \kappa (t) &= \dfrac{2}{(1 + 4t^2)^{\frac{3}{2}}} \end{align*} $$

Next, we find the critical points of the function by finding the zeros of the first derivative.

$$\begin{align*} \kappa (t) &= \dfrac{2}{(1 + 4t^2)^{\frac{3}{2}}} = 2(1 + 4t^2)^{-\frac{3}{2}} \\ \kappa'(t) &= 2(-\frac{3}{2})(1 + 4t^2)^{-\frac{5}{2}}(8t) \\ \kappa'(t) &= -24t(1 + 4t^2)^{-\frac{5}{2}} \\ 0 &= -24t(1 + 4t^2)^{-\frac{5}{2}} \\ 0 &= -24t \\ t &= 0 \end{align*} $$

The critical point of the function occurs at {eq}t = 0 {/eq}.

Since the domain of the function is {eq}(-\infty, \infty) {/eq} and we have already computed the first derivative, we use the first derivative test to classify the critical point as a maximum or minimum. First, we decompose the domain such that the critical point is excluded: {eq}(-\infty,0) \cup (0, \infty) {/eq}. Then, we select one value of {eq}x {/eq} from each interval and evaluate it using the first derivative. We select {eq}x = -1 {/eq} and {eq}x = 1 {/eq}. If the value of the first derivative is positive, the function is increasing over that entire interval. If the value of the first derivative is negative, the function is decreasing over that entire interval.

$$\begin{align*} \kappa'(-1) &= -24(-1)(1 + 4(-1)^2)^{-\frac{5}{2}} = 0.429 \\ \kappa'(1) &= -24(1)(1 + 4(1)^2)^{-\frac{5}{2}} = -0.429 \end{align*} $$

The function is increasing from {eq}(-\infty, 0) {/eq} and decreasing from {eq}(0, \infty) {/eq}. Therefore, the function has maximum curvature at {eq}x = 0 {/eq}.

To find the curvature at {eq}x = 0 {/eq}, we compute {eq}\kappa(0) {/eq}.

$$\kappa (0) = 2(1 + 4(0)^2)^{-\frac{3}{2}} = 2 $$

Therefore, the maximum curvature of the function is {eq}(0,2) {/eq}.

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from CAHSEE Math Exam: Tutoring Solution

Chapter 10 / Lesson 12