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### Part (a): Setting up the Integral to Determine the Arc Length
To find the length of the function [tex]\(f(x) = e^x \ln(x)\)[/tex] between [tex]\(x = 1\)[/tex] and [tex]\(x = 2\)[/tex], we use the arc length formula for a function [tex]\(y = f(x)\)[/tex].
The arc length [tex]\(L\)[/tex] of a curve [tex]\(y = f(x)\)[/tex] from [tex]\(x = a\)[/tex] to [tex]\(x = b\)[/tex] is given by:
[tex]\[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \][/tex]
Here, [tex]\(f(x) = e^x \ln(x)\)[/tex]. First, we need to compute the derivative [tex]\(\frac{dy}{dx}\)[/tex].
1. Calculate the Derivative:
Using the product rule for differentiation:
[tex]\[ f(x) = e^x \ln(x) \Rightarrow \frac{dy}{dx} = \frac{d}{dx}(e^x \ln(x)) \][/tex]
[tex]\[ \frac{dy}{dx} = \ln(x) \cdot \frac{d}{dx}(e^x) + e^x \cdot \frac{d}{dx}(\ln(x)) \][/tex]
[tex]\[ \frac{dy}{dx} = \ln(x) \cdot e^x + e^x \cdot \frac{1}{x} \][/tex]
[tex]\[ \frac{dy}{dx} = e^x \ln(x) + \frac{e^x}{x} \][/tex]
2. Substitute and Set Up the Integral:
Now, plug this derivative into the arc length formula:
[tex]\[ L = \int_1^2 \sqrt{1 + \left( e^x \ln(x) + \frac{e^x}{x} \right)^2} \, dx \][/tex]
So, the integral that represents the length of the function [tex]\(f(x) = e^x \ln(x)\)[/tex] from [tex]\(x = 1\)[/tex] to [tex]\(x = 2\)[/tex] is:
[tex]\[ \boxed{L = \int_1^2 \sqrt{1 + \left( e^x \ln(x) + \frac{e^x}{x} \right)^2} \, dx} \][/tex]
### Part (b): Converting Cartesian Coordinates to Polar Coordinates
To convert the Cartesian coordinates [tex]\((\sqrt{2}, -\sqrt{2})\)[/tex] to polar coordinates [tex]\((r, \theta)\)[/tex], we use the formulas:
[tex]\[ r = \sqrt{x^2 + y^2} \][/tex]
[tex]\[ \theta = \arctan\left(\frac{y}{x}\right) \][/tex]
Given [tex]\(x = \sqrt{2}\)[/tex] and [tex]\(y = -\sqrt{2}\)[/tex]:
1. Calculate [tex]\(r\)[/tex]:
[tex]\[ r = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} \][/tex]
[tex]\[ r = \sqrt{2 + 2} \][/tex]
[tex]\[ r = \sqrt{4} \][/tex]
[tex]\[ r = 2 \][/tex]
2. Calculate [tex]\(\theta\)[/tex]:
[tex]\[ \theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{-\sqrt{2}}{\sqrt{2}}\right) \][/tex]
[tex]\[ \theta = \arctan(-1) \][/tex]
The arctangent of [tex]\(-1\)[/tex] is [tex]\(-\frac{\pi}{4}\)[/tex]. However, this angle is not within the given range [tex]\(0 \leq \theta \leq 2\pi\)[/tex]. Since the point is in the fourth quadrant (positive [tex]\(x\)[/tex], negative [tex]\(y\)[/tex]), we need to add [tex]\(2\pi\)[/tex] to [tex]\(-\frac{\pi}{4}\)[/tex] to get it in the correct range:
[tex]\[ \theta = 2\pi - \frac{\pi}{4} = \frac{8\pi}{4} - \frac{\pi}{4} = \frac{7\pi}{4} \][/tex]
So, the polar coordinates [tex]\((r, \theta)\)[/tex] for the Cartesian coordinates [tex]\((\sqrt{2}, -\sqrt{2})\)[/tex] are:
[tex]\[ \boxed{(2, \frac{7\pi}{4})} \][/tex]
In summary:
1. The integral set-up for the arc length of [tex]\(f(x) = e^x \ln(x)\)[/tex] from [tex]\(x=1\)[/tex] to [tex]\(x=2\)[/tex] is:
[tex]\[ \int_1^2 \sqrt{1 + \left( e^x \ln(x) + \frac{e^x}{x} \right)^2} \, dx \][/tex]
2. The polar coordinates of [tex]\((\sqrt{2}, -\sqrt{2})\)[/tex] are:
[tex]\[ (2, \frac{7\pi}{4}) \][/tex]
To find the length of the function [tex]\(f(x) = e^x \ln(x)\)[/tex] between [tex]\(x = 1\)[/tex] and [tex]\(x = 2\)[/tex], we use the arc length formula for a function [tex]\(y = f(x)\)[/tex].
The arc length [tex]\(L\)[/tex] of a curve [tex]\(y = f(x)\)[/tex] from [tex]\(x = a\)[/tex] to [tex]\(x = b\)[/tex] is given by:
[tex]\[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \][/tex]
Here, [tex]\(f(x) = e^x \ln(x)\)[/tex]. First, we need to compute the derivative [tex]\(\frac{dy}{dx}\)[/tex].
1. Calculate the Derivative:
Using the product rule for differentiation:
[tex]\[ f(x) = e^x \ln(x) \Rightarrow \frac{dy}{dx} = \frac{d}{dx}(e^x \ln(x)) \][/tex]
[tex]\[ \frac{dy}{dx} = \ln(x) \cdot \frac{d}{dx}(e^x) + e^x \cdot \frac{d}{dx}(\ln(x)) \][/tex]
[tex]\[ \frac{dy}{dx} = \ln(x) \cdot e^x + e^x \cdot \frac{1}{x} \][/tex]
[tex]\[ \frac{dy}{dx} = e^x \ln(x) + \frac{e^x}{x} \][/tex]
2. Substitute and Set Up the Integral:
Now, plug this derivative into the arc length formula:
[tex]\[ L = \int_1^2 \sqrt{1 + \left( e^x \ln(x) + \frac{e^x}{x} \right)^2} \, dx \][/tex]
So, the integral that represents the length of the function [tex]\(f(x) = e^x \ln(x)\)[/tex] from [tex]\(x = 1\)[/tex] to [tex]\(x = 2\)[/tex] is:
[tex]\[ \boxed{L = \int_1^2 \sqrt{1 + \left( e^x \ln(x) + \frac{e^x}{x} \right)^2} \, dx} \][/tex]
### Part (b): Converting Cartesian Coordinates to Polar Coordinates
To convert the Cartesian coordinates [tex]\((\sqrt{2}, -\sqrt{2})\)[/tex] to polar coordinates [tex]\((r, \theta)\)[/tex], we use the formulas:
[tex]\[ r = \sqrt{x^2 + y^2} \][/tex]
[tex]\[ \theta = \arctan\left(\frac{y}{x}\right) \][/tex]
Given [tex]\(x = \sqrt{2}\)[/tex] and [tex]\(y = -\sqrt{2}\)[/tex]:
1. Calculate [tex]\(r\)[/tex]:
[tex]\[ r = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} \][/tex]
[tex]\[ r = \sqrt{2 + 2} \][/tex]
[tex]\[ r = \sqrt{4} \][/tex]
[tex]\[ r = 2 \][/tex]
2. Calculate [tex]\(\theta\)[/tex]:
[tex]\[ \theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{-\sqrt{2}}{\sqrt{2}}\right) \][/tex]
[tex]\[ \theta = \arctan(-1) \][/tex]
The arctangent of [tex]\(-1\)[/tex] is [tex]\(-\frac{\pi}{4}\)[/tex]. However, this angle is not within the given range [tex]\(0 \leq \theta \leq 2\pi\)[/tex]. Since the point is in the fourth quadrant (positive [tex]\(x\)[/tex], negative [tex]\(y\)[/tex]), we need to add [tex]\(2\pi\)[/tex] to [tex]\(-\frac{\pi}{4}\)[/tex] to get it in the correct range:
[tex]\[ \theta = 2\pi - \frac{\pi}{4} = \frac{8\pi}{4} - \frac{\pi}{4} = \frac{7\pi}{4} \][/tex]
So, the polar coordinates [tex]\((r, \theta)\)[/tex] for the Cartesian coordinates [tex]\((\sqrt{2}, -\sqrt{2})\)[/tex] are:
[tex]\[ \boxed{(2, \frac{7\pi}{4})} \][/tex]
In summary:
1. The integral set-up for the arc length of [tex]\(f(x) = e^x \ln(x)\)[/tex] from [tex]\(x=1\)[/tex] to [tex]\(x=2\)[/tex] is:
[tex]\[ \int_1^2 \sqrt{1 + \left( e^x \ln(x) + \frac{e^x}{x} \right)^2} \, dx \][/tex]
2. The polar coordinates of [tex]\((\sqrt{2}, -\sqrt{2})\)[/tex] are:
[tex]\[ (2, \frac{7\pi}{4}) \][/tex]
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