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1.3 1.3.1 Show, algebraically, that the graphs of \( y=2 \sin x \) and \( y=2 \cos x \) for \( 0 \leq x \leq \pi \) intersect at the point \( \left(\frac{\pi}{4} ; 1,414\right) \). Sketch the curves \( y=2 \sin x \) and \( y=2 \cos x \). Show the area bounded by the two curves and the \( y \)-axis. Show the representative strip/element that you will use to calculate the area bounded by the two curves, if the curves rotates about the \( x \)-axis.

Ask by Nunez Stewart. in South Africa
Nov 04,2024

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The graphs of \( y=2 \sin x \) and \( y=2 \cos x \) intersect at \( \left(\frac{\pi}{4}, 1.414\right) \). The area between the curves from \( x=0 \) to \( x=\frac{\pi}{4} \) is found by integrating \( 2 \cos x - 2 \sin x \). The volume of the solid formed by rotating this area about the \( x \)-axis is calculated using the integral \( \pi \int_0^{\frac{\pi}{4}} \left( (2 \cos x)^2 - (2 \sin x)^2 \right) \, dx \).

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To show that the graphs of \( y = 2 \sin x \) and \( y = 2 \cos x \) intersect at the point \( \left(\frac{\pi}{4}, 1.414\right) \), we need to set the two equations equal to each other: \[ 2 \sin x = 2 \cos x \] Dividing both sides by 2 gives: \[ \sin x = \cos x \] This equation holds true when \( x = \frac{\pi}{4} \) (or \( 45^\circ \)). To find the corresponding \( y \)-value: \[ y = 2 \sin\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414 \] Thus, \( (x, y) = \left(\frac{\pi}{4}, 1.414\right) \) confirms the intersection point. To sketch the curves \( y = 2 \sin x \) and \( y = 2 \cos x \) over the interval \( 0 \leq x \leq \pi \), note that \( y = 2 \sin x \) starts from 0, increases to 2 at \( x = \frac{\pi}{2} \), and then decreases back to 0 at \( x = \pi \). Meanwhile, \( y = 2 \cos x \) starts from 2, decreases to 0 at \( x = \frac{\pi}{2} \), and then decreases further to -2 at \( x = \pi \). The curves intersect at \( x = \frac{\pi}{4} \). To find the area bounded by the two curves and the \( y \)-axis, you can compute: \[ \text{Area} = \int_0^{\frac{\pi}{4}} (2 \cos x - 2 \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 \sin x - 2 \cos x) \, dx \] This splits the calculation into two regions based on the intersection. For the representative strip/element used to calculate the area bounded by the curves when they rotate around the \( x \)-axis, you would take a thin strip from the area between the functions at a position \( x \), where the height of the strip is given by the difference between the two curves, leading to cylindrical shell method or disk method application, depending on the exact setup you wish to employ. The radius of each shell or disk, in this case, corresponds to the \( y \)-values of the functions \( y = 2 \sin x \) or \( y = 2 \cos x \). Remember to sketch the area and curves appropriately for a visual understanding of their relationships and enclosed regions!

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