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Centent\newlineHome Pusa - C.\newlineCentent\newlinedance bla sean\newlineeellutd greent\newlineByatearn\newlinePress\newline(fin)F (fin) F \newlineto exit full screen\newlinedNJXCWUucXY11-GPfp22F33qWrTiclab1313a0606sp.\newlineTriognometric Functiont\newlineKABRIE\newlinesivetching the graph of \newliney=tan⁡(x) y=\tan(x) or \newliney=acot⁡(y) y=a \cot(y) \newlineEipahot\newlineGraph the trigonometric function.\newliney=−3cos⁡x y=-3\cos x \newlinePlot all points corresponding to \newlinex x -intercepts, minima, and maxima within one cycle. Then click on the graph-a-function button.\newline?

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Q. Centent\newlineHome Pusa - C.\newlineCentent\newlinedance bla sean\newlineeellutd greent\newlineByatearn\newlinePress\newline(fin)F (fin) F \newlineto exit full screen\newlinedNJXCWUucXY11-GPfp22F33qWrTiclab1313a0606sp.\newlineTriognometric Functiont\newlineKABRIE\newlinesivetching the graph of \newliney=tan⁡(x) y=\tan(x) or \newliney=acot⁡(y) y=a \cot(y) \newlineEipahot\newlineGraph the trigonometric function.\newliney=−3cos⁡x y=-3\cos x \newlinePlot all points corresponding to \newlinex x -intercepts, minima, and maxima within one cycle. Then click on the graph-a-function button.\newline?
  1. Understanding Cosine Function: To graph the function y=−3cos⁡(x)y = -3\cos(x), we need to understand the properties of the cosine function and how the coefficient −3-3 affects it.\newlineThe cosine function, cos⁡(x)\cos(x), has a period of 2π2\pi, meaning it repeats every 2π2\pi units. It has a maximum value of 11 and a minimum value of −1-1 when not scaled by any coefficient. The x-intercepts occur at π2\frac{\pi}{2} and 3π2\frac{3\pi}{2} within one cycle.\newlineThe coefficient −3-3 will stretch the graph vertically by a factor of −3-300 and reflect it across the x-axis because it is negative. This means the new maximum value will be −3-3 (since it's reflected) and the minimum value will be −3-300. The period remains 2π2\pi.
  2. Finding X-Intercepts: Let's find the x-intercepts of y=−3cos⁡(x)y = -3\cos(x). The x-intercepts occur where y=0y = 0.\newlineSetting y=−3cos⁡(x)y = -3\cos(x) to 00 gives us 0=−3cos⁡(x)0 = -3\cos(x). Dividing both sides by −3-3, we get cos⁡(x)=0\cos(x) = 0.\newlineThe solutions to cos⁡(x)=0\cos(x) = 0 within one cycle [0,2π][0, 2\pi] are x=π2x = \frac{\pi}{2} and y=0y = 000.
  3. Determining Minima and Maxima: Next, we find the minima and maxima. Since the function is −3cos⁡(x)-3\cos(x), the maxima occur where cos⁡(x)\cos(x) is at its minimum, which is −1-1, and the minima occur where cos⁡(x)\cos(x) is at its maximum, which is 11. For the maxima: y=−3cos⁡(x)=−3(−1)=3y = -3\cos(x) = -3(-1) = 3. This occurs at x=πx = \pi. For the minima: y=−3cos⁡(x)=−3(1)=−3y = -3\cos(x) = -3(1) = -3. This occurs at x=0x = 0 and x=2πx = 2\pi.
  4. Plotting Points and Drawing Curve: Now we can plot the points we have found on the graph:\newline- X-intercepts at (π2,0)(\frac{\pi}{2}, 0) and (3π2,0)(\frac{3\pi}{2}, 0).\newline- Maximum at (π,3)(\pi, 3).\newline- Minima at (0,−3)(0, -3) and (2π,−3)(2\pi, -3).\newlineWe can then draw the curve of the cosine function, keeping in mind the vertical stretch and reflection across the x-axis.

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Rakesh tried to solve the differential equation dydx=2y⋅cos⁡(x) \frac{d y}{d x}=2 y \cdot \cos (x) . This is his work:\newlinedydx=2y⋅cos⁡(x) \frac{d y}{d x}=2 y \cdot \cos (x) \newlineStep 11: ∫1ydy=∫2cos⁡(x)dx \quad \int \frac{1}{y} d y=\int 2 \cos (x) d x \newlineStep 22: ln⁡∣y∣=2sin⁡(x) \quad \ln |y|=2 \sin (x) \newlineStep 33: eln⁡∣y∣=e2sin⁡(x) \quad e^{\ln |y|}=e^{2 \sin (x)} \newlineStep 44: ∣y∣=e2sin⁡(x) \quad|y|=e^{2 \sin (x)} \newlineStep 55: y=±e2sin⁡(x)+C \quad y= \pm e^{2 \sin (x)}+C \newlineIs Rakesh's work correct? If not, what is his mistake?\newlineChoose 11 answer:\newline(A) Rakesh's work is correct.\newline(B) Step 11 is incorrect. The separation of variables wasn't done correctly.\newline(C) Step 22 is incorrect. The right-hand side of the equation should be 2sin⁡(x)+C 2 \sin (x)+C .\newline(D) Step 44 is incorrect. eln⁡∣y∣ e^{\ln |y|} isn't equal to ∣y∣ |y| .
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Posted 3 months ago

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Posted 3 months ago

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Posted 3 months ago

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Water is being heated in a kettle. At time tt seconds, the temperature of the water is θ∘C\theta^{\circ}C. The rate of increase of the temperature of the water at time tt is modelled by the differential equation\newlinedθdt=λ(120−θ)θ≤100\frac{d\theta}{dt}=\lambda(120-\theta)\quad \theta \leq 100\newlinewhere λ\lambda is a positive constant. Given that θ=20\theta=20 when t=0t=0\newline(a) solve this differential equation to show that\newlineθ=120−100e−λt\theta=120-100e^{-\lambda t}\newline(88)
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Posted 3 months ago

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Posted 2 months ago

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Posted 2 months ago

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Posted 2 months ago

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Posted 2 months ago

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Posted 2 months ago

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