Consider the function. y = ln (3x) Do the following. a) At what point does the curve have maximum...

Question:

Consider the function.

y = ln (3x)

Do the following.

a) At what point does the curve have maximum curvature?

b) What happens to the curvature as {eq}x\rightarrow \infty {/eq}?

Maximum Curvature:

To find the maximum curvature of a function, {eq}f(x) {/eq}, we must first find the curvature function, {eq}\kappa (x) = \frac{|f''(x)|}{(1+ (f'(x))^2)^{\frac{3}{2}}} {/eq}. Then, we find the critical points of the curvature function by differentiating {eq}\kappa(x) {/eq}, setting it equal to {eq}0 {/eq} and solving for {eq}x {/eq}. Then, we employ the first derivative test to classify the obtained values of {eq}x {/eq} as maximums or minimums. In the first derivative test, we sample values of {eq}x {/eq} to the left and right of the critical points, evaluate them using the first derivative. If the value of the first derivative is positive, the function is increasing over that interval. If the first derivative is negative, the function is decreasing over that interval.

Answer and Explanation:

We must first find the curvature function. To do this, we must find the first and second derivatives of the original function.

$$\begin{align*} f(x) &= \ln(3x) \\ f'(x) &= \frac{1}{x} \\ f''(x) &= -\frac{1}{x^2} \end{align*} $$

Then, we generate the curvature function and simplify.

$$\begin{align*} \kappa (x) &= \dfrac{|f''(x)|}{(1 + (f'(x))^2)^{\frac{3}{2}}} \\ \kappa (x) &= \dfrac{|-\frac{1}{x^2}|}{(1 + (\frac{1}{x})^2)^{\frac{3}{2}}} \\ \kappa (x) &= \dfrac{\frac{1}{x^2}}{(1 + \frac{1}{x^2})^{\frac{3}{2}}} \\ \kappa (x) &= \dfrac{1}{x^2(1 + \frac{1}{x^2})^{\frac{3}{2}}} \\ \kappa (x) &= \dfrac{1}{x^2(x^2 + 1)^{\frac{3}{2}}} \\ \end{align*} $$

a) At what point does the curve have maximum curvature?

To find the point of maximum curvature, we must first find the critical points of the curvature function. To do this, we find the first derivative of the function, set it equal to {eq}0 {/eq} and solve for {eq}x {/eq}.

$$\begin{align*} \kappa (x) &= \dfrac{1}{x^2(x^2 + 1)^{\frac{3}{2}}} \\ \kappa'(x) &= \dfrac{x^2(x^2 + 1)^{\frac{3}{2}}(0) - (1)[x^2(\frac{3}{2})(x^2 + 1)^{\frac{1}{2}}(2x)]}{(x^2(x^2 + 1)^{\frac{3}{2}})^2} \\ \kappa'(x) &= \dfrac{-3x^3(x^2 + 1)^{\frac{1}{2}}}{x^4(x^2 + 1)^3} \\ 0 &= \dfrac{-3x^3\sqrt{x^2 + 1}}{x^4(x^2 + 1)^3} \\ 0 &= -3x^3\sqrt{x^2 + 1} \\ x &= 0 \\ 0 &= x^4(x^2 + 1)^3 \\ x &= 0 \end{align*} $$

Since {eq}x^2 + 1 \neq 0 {/eq}, we obtain only the critical point {eq}x = 0 {/eq}.

To ensure we have identified the point of maximum curvature, we apply the first derivative test. We decompose the domain of the curvature function such that the critical point is excluded. This yields {eq}(-\infty, 0) \cup (0, \infty) {/eq}. Then, we select one value of {eq}x {/eq} from each interval and evaluate it using {eq}\kappa'(x) {/eq}. We select the values {eq}x = -1 {/eq} and {eq}x = 1 {/eq}.

$$\begin{align*} \kappa'(-1) &= \dfrac{-3(-1)^3\sqrt{(-1)^2+1}}{(-1)^4((-1)^2 + 1)^3} = \frac{3\sqrt2}{8} \\ \kappa'(1) &= \dfrac{-3(1)^3\sqrt{(1)^2 + 1}}{(1)^4((1)^2 + 1)^3} = -\frac{3\sqrt2}{8} \end{align*} $$

Based on the values of {eq}kappa' {/eq}, we determine the function is increasing over the interval {eq}(-\infty,0) {/eq} and decreasing over the interval {eq}(0, \infty) {/eq}. Therefore, the function has maximum curvature at {eq}x = 0 {/eq}.

b). What happens to the curvature as {eq}x \rightarrow \infty {/eq}?

Since the curvature of the function is decreasing from {eq}(0, \infty) {/eq} based on the results of the first derivative test, we can conclude that the curvature of the function is decreasing as {eq}x \rightarrow \infty {/eq}.


Learn more about this topic:

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Using Differentiation to Find Maximum and Minimum Values

from Math 104: Calculus

Chapter 9 / Lesson 4
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