Curve Length Ad4670

1. **Problem Statement:** Find a curve $y = f(x)$ passing through the point $(1,1)$ with a positive derivative $f'(x) > 0$ such that the length of the curve from $x=1$ to $x=4$ is given by $$L = \int_1^4 \sqrt{1 + \frac{1}{4x}} \, dx.$$ 2. **Formula for Arc Length:** The length $L$ of a curve $y=f(x)$ from $x=a$ to $x=b$ is given by $$L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx.$$ 3. **Matching the given integral:** We have $$\sqrt{1 + (f'(x))^2} = \sqrt{1 + \frac{1}{4x}}.$$ Squaring both sides, $$1 + (f'(x))^2 = 1 + \frac{1}{4x}.$$ 4. **Solve for $f'(x)$:** $$(f'(x))^2 = \frac{1}{4x}.$$ Taking the positive root (since derivative is positive), $$f'(x) = \frac{1}{2\sqrt{x}}.$$ 5. **Find $f(x)$ by integration:** $$f(x) = \int f'(x) \, dx = \int \frac{1}{2\sqrt{x}} \, dx = \int \frac{1}{2} x^{-\frac{1}{2}} \, dx.$$ Integrate: $$f(x) = \frac{1}{2} \cdot \frac{2}{1} x^{\frac{1}{2}} + C = \sqrt{x} + C.$$ 6. **Use initial condition $f(1) = 1$ to find $C$:** $$1 = \sqrt{1} + C \implies C = 0.$$ 7. **Final curve:** $$f(x) = \sqrt{x}.$$ 8. **Answer to part b:** Since the arc length formula depends on $(f'(x))^2$, the derivative $f'(x)$ could be either positive or negative root. But the problem requires a positive derivative, so only one such curve exists. **Summary:** - The curve is $f(x) = \sqrt{x}$. - There is exactly one such curve with positive derivative passing through $(1,1)$ that satisfies the given arc length integral.