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Tutor profile: Lilly S.

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Lilly S.
Mathematics Major and Freelance Tutor for 4 years
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Questions

Subject: Physics (Newtonian Mechanics)

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Question:

At what velocity do I need to spin a 10kg bucket of water vertically on a 1 meter rope such that no water falls out of the bucket?

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Lilly S.
Answer:

The centripetal force is not a real force strictly speaking, but it is easier to think about this problem in terms of it. For the water to stay in the bucket when it's at the top of its swing, the centripetal force on the water needs to be equal to the force of gravity pulling it downwards. The force of gravity is just 9.8 multiplied by the mass of the water, 10kg, or 98N. Since our unknown is the velocity, we can let v be a variable representing the velocity of the water and find the centripetal force in terms of v. The centripetal force is mass * velocity^2 / the radius of the circle. Filling in 10kg for the mass and 1m for the radius of the circle, we get that the centripetal force = 10v^2. Thus, since we need the force of gravity and the centripetal force on the water to be equal, we can set these two equal and solve for v. (10kg)v^2 = 98N --> v^2 = 9.8 m^2/s^2 --> v = 3.13 m/s. The speed you need to spin a 10kg bucket of water on a 1 meter rope to keep the water from splashing out is 3.13 m/s.

Subject: Linear Algebra

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Question:

What are the eigenvectors of a 2x2 rotation matrix, A? What about its determinant?

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Lilly S.
Answer:

Since I didn't specify which rotation matrix, we must assume that A is a general 2x2 rotation matrix. Well we could write this out explicitly and then attempt to solve for its eigenvalues, then use those to find its eigenvectors, we can avoid all that tedious calculation by instead asking: what are eigenvectors? An eigenvector is a vector such that applying (multiplying on the left) the matrix A to that eigenvector only multiplies the original vector by some number. That number is known as the eigenvector's eigenvalue. Or in other words, an eigenvector is a vector that stays on the same line as the matrix A transforms the plane. Notice, if A represents any 2D rotation of the plane except $$\pi$$ or $$2\pi$$, it will by definition rotate all vectors off of their original line. Thus, almost all 2x2 rotation matrices have no eigenvectors at all. If A represents a rotation of $$\pi$$ radians, then all vectors are eigenvectors, since the effect of A is to reverse the direction of all vectors. In fact, all these eigenvectors have an eigenvalue of -1. If A represents a rotation of $$2\pi$$ radians, this is actually the identity transformation and thus all vectors in the plane of eigenvectors with an eigenvalue of 1. Next, what is the determinant of A? Again, we could write out the general form and go through the process of computing |A|, but there is a simpler way to think about it. The determinant of a matrix describes how that matrix squishes or shrinks the plane. But a rotation doesn't squish or shrink the plane at all. Thus, we know that the determinant of any 2x2 rotation matrix must be 1, that is, areas stay the same under a rotation.

Subject: Calculus

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Question:

What is the integral of $$\arctan(x)$$?

Inactive
Lilly S.
Answer:

To compute this integral, we can first use the technique of trig sub. Trig sub is often useful for inverse trig functions since it replaces them with the more familiar trig functions. In this example, we can let $$x = \tan(u)$$. Thus, $$\arctan(x) = \arctan(\tan(u)) = u$$, since arctan and tan are inverses of each other. Taking the derivative of both sides, $$dx = \sec^2(u)du$$. Thus, the integral becomes $$\int{u\cdot{}\sec^2(u)du}$$. Now, we have the integral of a product, which means we can use Integration by Parts. Differentiating $$u$$, we get $$1$$, then $$0$$. Integrating $$sec^2(u)$$ we get $$tan(u)$$. To find the integral of $$tan(u)$$, we can use trigonometry to rewrite it as $$\frac{\sin{u}}{\cos{u}}$$. Now, we let $$t=\cos{u}$$, and thus $$dt=-\sin{u}du$$. Using this substitution, $$\int{\tan(u)du}=\int{\frac{\sin{u}}{\cos{u}}du}=-\int{\frac{dt}{t}}=-\ln|t|=-\ln|\cos{u}|$$. Since $$-\ln(x)=\ln(\frac{1}{x})$$, the integral of $$\tan(u)$$ is $$\ln|\frac{1}{\cos{u}}|=\ln|\sec{u}|$$. So, using Integration by Parts, we get that $$\int{u\cdot\sec^2(u)du}=u\cdot\tan(u)-1\cdot\ln|\sec(u)|+\int{0\cdot\ln|\sec(u)|}du=u\tan(u)-\ln|\sec(u)|$$. To get things in term of $$x$$, we use our substitution from before: $$u=\arctan(x)$$. Thus, $$\int{\arctan(x)}dx=\arctan(x)\cdot\tan(\arctan(x))-\ln|\sec(\arctan(x))|$$, which equals $$x\arctan(x)-\ln\sqrt{1+x^2}$$.

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