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The simplified form of the function h(t)=∫ᵗ₀ (10cos40t+40sin40t) at is ______ cos(40t+(_____) Please report yout answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 180 degrees.

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Final answer:

The simplified form of the function \( h(t) \) is \( \frac{\sqrt{17}}{4}\cos(40t + \arctan(-\frac{1}{4})) \), where the cosine has a positive magnitude and the angle is within the range of -180 to 180 degrees.

Step-by-step explanation:

The simplified form of the function \( h(t)=\int_{t_0}^{t} (10\cos(40t)+40\sin(40t)) \, dt \) can be found by integrating the function with respect to \(t\). When we integrate \(10\cos(40t)\), we get \(\frac{10}{40}\sin(40t)\), and when we integrate \(40\sin(40t)\), we get \(-\frac{40}{40}\cos(40t)\) which simplifies to \(-\cos(40t)\). Therefore, the integral of the given function is \(\frac{1}{4}\sin(40t) - \cos(40t) + C\), where \(C\) is the constant of integration. However, this is not in the desired form of a cosine function with a positive magnitude.

To convert this result into a single cosine function with a phase shift, we use the identity \(a\cos(x) + b\sin(x) = r\cos(x - \phi)\) where \(r = \sqrt{a^2 + b^2}\) and \(\phi = \arctan(\frac{b}{a})\). In our case, \(a = -1\) and \(b = \frac{1}{4}\), which gives us \(r = \sqrt{(-1)^2 + (\frac{1}{4})^2} = \sqrt{1 + \frac{1}{16}} = \sqrt{\frac{17}{16}} = \frac{\sqrt{17}}{4}\), and \(\phi = \arctan(-\frac{1}{4})\). The \(\arctan\) function gives an angle in the fourth quadrant, so we adjust it to find an angle between -180 degrees and 180 degrees.

Thus, the simplified form of the function is \( h(t) = \frac{\sqrt{17}}{4}\cos(40t + \arctan(-\frac{1}{4}))\) with a positive magnitude and an angle within the desired range.

User Ilyas Patanam
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