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Tutor profile: Vishal B.

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Vishal B.
Mathematician and Graduate Teaching Assistant (for 3+ years)
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Questions

Subject: Applied Mathematics

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Question:

Suppose a random variable X follows an exponential distribution; thats is, X ~ exp($$\lambda$$). Let $$Y=X^2$$ be another random variable. Find the pdf of Y.

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Vishal B.
Answer:

To compute the pdf of Y, we first compute its cdf. Since X is exponentially distributed, it is easier to compute the survival function, and then compute its cdf because cdf = 1 - survival function. $$P(Y>y)=P(X^2>y)=P(X>y^\frac{1}{2})=e^{-\lambda*y^\frac{1}{2}}$$ $$P(Y<=y) = 1 - e^{-\lambda*y^\frac{1}{2}}$$ pdf of Y = derivative of the cdf of Y with respect to y pdf of y = $$\frac {dP(Y<=y)}{dy}$$ pdf of Y = $$\frac {\lambda e^{-\lambda y^\frac{1}{2}}} {2y^\frac{1}{2}}$$

Subject: Calculus

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Question:

Let $$f(x) = \frac{1}{x} $$ on $$0<x<=1$$, and $$f(x)=0$$ at $$x=0$$. Is f(x) continuous on [0,1]? Give your reasoning.

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Vishal B.
Answer:

The function f is not continuous on [0,1]. $$f(x) = \frac{1}{x} $$ We know that f is continuous on the following interval (0,1] It remains to check whether f is continuous at 0. For f to be continuous at x=0 the following conditions must be satisfied: 1. f(0) is well-defined 2. $$\lim_{x \to 0} f(x)$$ exists 3. $$\lim_{x \to 0} f(x) = f(0)$$ Let us check whether the above conditions are satisfied. 1. $$f(0)=0$$ by assumption. So condition 1 is satisfied. 2. $$\lim_{x \to 0} f(x) = \lim_{x \to 0}\frac{1}{x} = \infty$$. So condition 2 is satisfied. 3. However, condition 3 is clearly not satisfied; thats is, $$\lim_{x \to 0} f(x) != f(0)$$ Thus, f is not continuous at x=0, hence, f is not continuous on the interval [0,1].

Subject: Physics

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Question:

Suppose you have two objects of equal mass: a ball and a box. You place the two object at the top of an inclined plane. Assume there is no friction between the box and the surface of the inclined plane, and that the ball rolls as it should. Gravity is the only external force that the objects experience. At the end of the inclined place, how will the velocity of the ball compare to the velocity of the box?

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Vishal B.
Answer:

The velocity of the box will be greater than the velocity of the ball. Note: The beauty of the problems is that we can figure out the solution without explicit computations! Let us consider the following thought experiment. Both the objects start at the same height above the ground; that is, at the top of the inclined plane. Since they both have the same mass (by assumption), they both have the same initial potential energy. Now when they descend down the inclined plane, their potential energy is converted into kinetic energy. For this problem, we consider the two different kinds of kinetic energy, namely, translational kinetic energy and rotational kinetic energy. Since the box is sliding down the slope (without friction), all the potential energy is converted into translational kinetic energy; whereas, the potential energy of the ball is converted into translational and rotational kinetic energy. Now we apply the law of conservation of energy to the problem. For the box: $$Initial potential energy of the box = Final translational kinetic energy of the box$$ For the ball: $$Initial potential energy of the ball = Final translational kinetic energy of the ball + final rotational kinetic energy of the ball$$ This implies: $$Final translational kinetic energy of the ball = Initial potential energy of the ball - final rotational kinetic energy of the ball < Initial potential energy of the ball = initial potential energy of the box$$ Note: Translation kinetic energy = $$0.5*mass*velocity^2$$ Therefore, we can conclude that the velocity of the box is greater than the velocity of the ball.

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