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Tutor profile: Steven U.

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Steven U.
Graduate with B.A. in Mathematics; All Levels of Calculus Welcome
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Questions

Subject:Linear Algebra

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Question:

Let R denote the set of all real numbers. Consider the function f: R-->R given by f(x)=mx+b, where b is nonzero. Is f a linear transformation from R to R? Why or why not?

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Steven U.

If f is to be a linear transformation from R to R, by definition f(0) must be zero. But f(0)=b, which is nonzero. So f is not a linear transformation. Observe that there is a possibility for confusion with the equation for a line in the xy-plane "y=mx+b", which is often called a linear equation, or a linear function. However, a linear function, or linear transformation, in the linear-algebraic sense is not the same.

Subject:Trigonometry

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Question:

Suppose someone draws an angle on a piece of paper, and asks you to measure it in degrees, without a protractor. You do have a rigid ruler and a pencil, though. What is a mathematically-sound way to measure the angle?

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Steven U.

Here is one answer: ________ Place the ruler in the plane of the paper, and trace a fixed length along each of the angle's two rays, starting from the vertex of the angle, and mark the points where you stop, one for each ray. Both lengths should be equal. Assign this common length to be 1 unit . Then draw a circular arc whose central angle is the angle you are trying to measure, connecting the two endpoints you marked just before. Now, as best as you can, observe the length of this circular arc, and determine what fraction of the unit length this arc length is. (For example, 2/3 of the unit length, or 6/5 of the unit length, etc.). Then multiply this fraction by 180/3.14, and the resulting number is at least a rough but decent estimate of the angle's measure in degrees. ________ Explanation: The justification of this answer is the following. Recall that the length s of the circular arc of radius r with central angle of measure $\theta$ radians is $s=(r)(\theta)$. Therefore, in radians, $\theta= frac{s}{r}$. This is the same set-up we have in this question. In the answer, we took r=1 to avoid the possible error associated with dividing the arc length s by non-unit r. Thus the angle measure in radians is the ratio of our arc length to our radius, which is the arc length itself, because we took radius r=1. The resulting angle is in radians; we must multiply the arc lengthby 180/$\pi$ to obtain a degree measure. Of course, we used the approximation 3.14 for $\pi$.

Subject:Calculus

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Question:

Evaluate the indefinite integral $\int\sin^{3}{t}dt$.

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Steven U.

To evaluate this integral, we will use our trigonometric identities (these are useful for at least several other types of integrands you encounter in calculus). Use the Pythagorean identity $\sin^{2}{t}+\cos^{2}{t}=1$, slightly rearranged, to write the integrand as $\sin^{3}{t}=\sin{t}\sin^{2}{t}=\sin{t}(1-\cos^{2}{t})=\sin{t}-\sin{t}\cos^{2}{t}$. So our integral is $$\int\sin^{3}{t}dt=\int(\sin{t}-\sin{t}\cos^{2}{t})dt$$ which, by our basic properties of indefinite integrals (namely, here, that the integral of a sum is the sum of integrals) is $$\int\sin^{3}{t}dt=\int\sin{t}dt-\int\sin{t}\cos^{2}{t}dt.$$ Recalling that the derivative of $-\cos{t}$ with respect to $t$ is $\sin{t}$ (that is,$\frac{d}{dt}(-\cos{t})=-\frac{d}{dt}\cos{t}=-(-\sin{t})=\sin{t}$), we immediately have $\int\sin{t}dt=-\cos{t}+C$ ,where $C$ is an arbitrary constant (usually called the constant of integration). To evaluate the second integral $$\int\sin{t}\cos^{2}{t}dt,$$ we notice that the derivative of $\cos{t}$ with respect to $t$ is $-\sin{t}$, and we have a factor of $\sin{t}$ in the integrand, so we may change variables. Define a new variable $u$ by $u=\cos{t}$, so that $\frac{du}{dt}=-\sin{t}$, from which we have the formal expression $du=-\sin{t}dt$ The integral contains the factor $\sin{t}dt$, so we multiply the formal expression by $-1$ to get $-du=\sin{t}dt$. We now substitute all our expressions in $t$ with the equal expressions in $u$ to rewrite the integral as $$\int\sin{t}\cos^{2}{t}dt=\int-u^2du$$. We immediately use the power rule to obtain $$\int\sin{t}\cos^{2}{t}dt=\int-u^2du=-\frac{u^3}{3}+D=-\frac{\cos^{3}{t}}{3}+D,$$ where $D$ is our constant of integration and in the last equality we substituted back the original expressions in $t$. The final step is to synthesize our results by writing $$\int\sin^{3}{t}dt=\int\sin{t}dt- \int\sin{t}\cos^{2}{t}dt=-\cos{t}-(-\frac{\cos^{3}{t}}{3})+M=-\cos{t}+\frac{\cos^{3}{t}}{3}+M$$, where $M=C+D$ is no more than a just new arbitrary constant of integration, and we are done.

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