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Math  /  Calculus

QuestionAt what point does the curve have maximum curvature? y=3ex(x,y)=(□)\begin{array}{c} y=3 e^{x} \\ (x, y)=(\square) \end{array}
What happens to the curvature as x→∞x \rightarrow \infty ? x(x)x(x) approaches □\square as x→∞x \rightarrow \infty,

Studdy Solution

STEP 1

1. The curvature of a curve given by y=f(x) y = f(x) can be calculated using the formula for curvature κ(x) \kappa(x) .
2. The formula for curvature κ(x) \kappa(x) of a function y=f(x) y = f(x) is given by: $ \kappa(x) = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}} \]
3. We need to find the point where this curvature is maximized.
4. We also need to analyze the behavior of the curvature as \( x \rightarrow \infty \).

STEP 2

1. Calculate the first and second derivatives of y=3ex y = 3e^x .
2. Substitute these derivatives into the curvature formula.
3. Find the value of x x that maximizes the curvature.
4. Determine the behavior of the curvature as x→∞ x \rightarrow \infty .

STEP 3

Calculate the first derivative f′(x) f'(x) of y=3ex y = 3e^x :
f′(x)=ddx(3ex)=3ex f'(x) = \frac{d}{dx}(3e^x) = 3e^x

STEP 4

Calculate the second derivative f′′(x) f''(x) :
f′′(x)=ddx(3ex)=3ex f''(x) = \frac{d}{dx}(3e^x) = 3e^x

STEP 5

Substitute f′(x) f'(x) and f′′(x) f''(x) into the curvature formula:
κ(x)=∣3ex∣(1+(3ex)2)3/2 \kappa(x) = \frac{|3e^x|}{(1 + (3e^x)^2)^{3/2}}
Since 3ex 3e^x is always positive, we have:
κ(x)=3ex(1+9e2x)3/2 \kappa(x) = \frac{3e^x}{(1 + 9e^{2x})^{3/2}}

STEP 6

To find the maximum curvature, we need to find the critical points of κ(x) \kappa(x) . This involves taking the derivative of κ(x) \kappa(x) with respect to x x and setting it to zero. However, due to the complexity of the expression, we can use a qualitative analysis to determine the behavior.
Notice that as x→−∞ x \rightarrow -\infty , ex→0 e^x \rightarrow 0 , so κ(x)→0 \kappa(x) \rightarrow 0 .
As x→∞ x \rightarrow \infty , ex→∞ e^x \rightarrow \infty , and κ(x)→0 \kappa(x) \rightarrow 0 .
The maximum curvature occurs at some finite x x .

STEP 7

Analyze the behavior of the curvature as x→∞ x \rightarrow \infty :
As x→∞ x \rightarrow \infty , ex→∞ e^x \rightarrow \infty , making the denominator (1+9e2x)3/2 (1 + 9e^{2x})^{3/2} grow much faster than the numerator 3ex 3e^x . Therefore, κ(x)→0 \kappa(x) \rightarrow 0 .
The point of maximum curvature is not explicitly calculated here due to the complexity, but it occurs at a finite x x . As x→∞ x \rightarrow \infty , the curvature κ(x)→0 \kappa(x) \rightarrow 0 .

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