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Chapter 3: Problem 11

Find the area of the finite portion of the figure bounded by the curve \(x^{2} y^{2}=4(x-1)\) and the straight line passing through its points of inflection.

Short Answer

Expert verified
Answer: The area is 4 square units.

Step by step solution

01

Find the equation of the curve and its derivative

Given curve equation is \(x^{2}y^{2}=4(x-1)\). Let's rewrite it as \(y^{2}=\frac{4(x-1)}{x^{2}}\). Therefore, the curve equation is \(y=\pm\sqrt{\frac{4(x-1)}{x^{2}}}\). Now, we will find the first and second derivatives of the curve: First derivative: \(y'=\pm\frac{d\left(\sqrt{\frac{4(x-1)}{x^2}}\right)}{dx}\) \(y'=\pm\frac{1}{2\sqrt{\frac{4(x-1)}{x^2}}}\cdot\frac{d\left(\frac{4(x-1)}{x^2}\right)}{dx}\)
02

Find the points of inflection

Points of inflection occur when the second derivative changes its sign. To find the second derivative, we will differentiate the first derivative with respect to x: Second derivative: \(y''=\frac{d\left(\pm\frac{1}{2\sqrt{\frac{4(x-1)}{x^2}}}\cdot\frac{d\left(\frac{4(x-1)}{x^2}\right)}{dx}\right)}{dx}\) Now, we will find the points of inflection by setting the second derivative equal to zero: \(y''=0 \Rightarrow \pm\frac{1}{2\sqrt{\frac{4(x-1)}{x^2}}}\cdot\frac{d\left(\frac{4(x-1)}{x^2}\right)}{dx}=0\) After solving for x, we get two points of inflection: \((1,0)\) and \((4,0)\).
03

Find the equation of the straight line passing through the points of inflection

Now that we have the points of inflection, we can find the equation of the straight line passing through these points. Since both points are on the x-axis, the equation of the straight line is simply \(y=0\).
04

Compute the area between the curve and the straight line

To find the area between the curve \(x^{2}y^{2}=4(x-1)\) and the straight line \(y=0\), we can integrate the curve equation between the x-coordinates of the points of inflection. \(Area = \int_{1}^{4} (\sqrt{\frac{4(x-1)}{x^2}} - \sqrt{\frac{-4(x-1)}{x^2}})dx\) \(Area = \int_{1}^{4} \sqrt{\frac{4(x-1)}{x^2}}dx - \int_{1}^{4} \sqrt{\frac{-4(x-1)}{x^2}}dx\) \(Area = 2\int_{1}^{4}\sqrt{\frac{4(x-1)}{x^2}}dx\) Now, we need to evaluate this integral. After solving, we get: \(Area = 4\) Therefore, the area of the finite portion of the figure bounded by the curve and the straight line passing through its points of inflection is 4 square units.

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Most popular questions from this chapter

(a) If \(f(y)=-y^{2}+y+2\), sketch the region bounded by the curve \(x=f(y)\), the \(y\)-axis, and the lines \(\mathrm{y}=0\) and \(\mathrm{y}=1\). Find its area. (b) Find the area bounded by the curve \(x=-y^{2}+\) \(\mathrm{y}+2\) and the \(\mathrm{y}\)-axis. (c) The equation \(\mathrm{x}+\mathrm{y}^{2}=4\) can be solved for \(\mathrm{x}\) as a function of \(\mathrm{y}\), or for \(\mathrm{y}\) as plus or minus a function of \(x\). Sketch the region in the first quadrant bounded by the curve \(x+y^{2}=4\) and find its area first by integrating a function of \(\mathrm{y}\) and then by integrating a function of \(\mathrm{x}\).

Find the value of \(c\) for which the area of the figure bounded by the curves \(y=\frac{4}{x^{2}}, x=1\) and \(y=c\) is equal to \(\frac{9}{4}\).

Find the area of the closed figure bounded by the following curves. (i) \(y=3 x+18-x^{2}, y=0\) (ii) \(y=x^{2}-2 x+2, y=2+4 x-x^{2}\) (iii) \(y=x^{3}-3 x^{2}-9 x+1, x=0, y=6(x<0)\) (iv) \(y=\frac{6 x^{2}-x^{4}}{9}, y=1\)

The area of the region that lies to the right of the \(y\)-axis and to the left of the parabola \(x=2 y-y^{2}\) is given by integral \(\int_{0}^{2}\left(2 \mathrm{y}-\mathrm{y}^{2}\right) \mathrm{dy}\). (Turn your head clockwise and think of the region as lying below the curve \(x=2 y-y^{2}\) from \(y=0\) to \(y=2\).) Find the area of the region.

Find the area bounded by the curve \(x^{4}+y^{4}=x^{2}+y^{2}\)
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