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# Tutor profile: Smiti N.

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Smiti N.
Engineering Student at UCLA
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## Questions

### Subject:Calculus

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Question:

What are related rates and how do you approach a related rate problem?

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Smiti N.

A related rate is when the rate of change of one quantity depends on the rate of change of another quantity. For example, if the length of the sides of a square is changing with time, the area of the square will also change with time. The rate of change o the area depends on the rate of change of the side length. The first step to tackling a related rate problem is to determine which two rates are related and how. Let us break down how to solve the following practical example: There is an empty pool with a length of 50 m, width of 25 m, and height of 2 m. Suppose the pool is being filled at a rate of $$10 \frac{m^3}{minute}$$. At what rate is the depth of the water increasing? How long will it take for the pool to be filled? $$\textbf{Step 1:}$$ Determine the two related rates The rate given in the problem is the rate at which the pool is filled (1000 L/min). This is the rate of change of the volume (V) . The second rate is the one asked in the question - the rate of change of the water depth (h). Therefore our two related rates are $$\frac {dV}{dt}$$ and $$\frac {dh}{dt}$$ $$\textbf{Step 2:}$$ Determine how the two quantities are related The two quantities are volume (V) and depth (h). The pool is a rectangular prism, so the equation for the volume is $$V = l \times w \times h$$ $$h = \frac{V}{l \times w}$$ where $$l$$ and $$w$$ are the length and width, respectively. $$\textbf{Step 3:}$$ Determine how the two quantities are related To do this, we take the derivative of the equation we found in the previous step. Before we do this, recall that the length and width of the pool are not changing and are constants. Therefore, we treat them as constants when taking the derivative. $$\frac{dh}{dt} = \frac{dV}{dt} \cdot \frac{1}{l \times w}$$ $$\textbf{Step 4:}$$ Substitute the known rate into the equation and solve $$\frac{dh}{dt} = 10 \frac{m^3}{minute} \cdot \frac{1}{50 m \times 25 m}$$ $$\frac{dh}{dt} = \frac{10}{50 m \times 25 m} \frac{m^3}{minute}$$ $$\frac{dh}{dt} = 0.008 \frac{m}{minute}$$ Now we have solved the first part of the question using the 4 steps of basic related rate problems. We use the rate of the change in depth to solve for the second part. $$\frac{dh}{dt} = 0.008 \frac{m}{minute}$$ Let $$\alpha$$ be the number of minutes it takes for the depth of the pool to reach 2 m. $$2 m = \frac{dh}{dt} \times \alpha$$ $$\alpha = \frac {2 m}{\frac{dh}{dt}}$$ $$\alpha = \frac {2 m}{0.008 \frac{m}{minute}}$$ $$\alpha = 250 minutes$$

### Subject:Physics

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Question:

What is the relationship between work and power?

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Smiti N.

Work is done on an object when a force applied to the object causes it to move. The key thing here is that the object has to move. For example, if you push against a wall and it stays put, work was NOT done on the object even though you applied a force to it. But if you were pushing a box from one end of a room to the other, work was done on the box because the force of your pushing caused it to move. $$Work = Force \times distance\times cos(\theta)$$ where $$\theta$$ is the angle between the force vector and the direction of displacement. Work is measured in the standard unit Joules (J). $$1 Joule = 1 N\cdot m$$ Note that work is not time dependent. If you pushed the box 5 meters in 10 minutes versus in 1 hour, the same amount of work would have been done on the box because the distance it travelled was the same. Power, however, is time dependent. Power is the rate at which work is done. $$Power = \frac{work}{time}$$ The standard unit of power is Watts (W). $$1 Watt = 1 J/s$$ To demonstrate this concept, let us consider the box example. Suppose you are applying a $$120 N$$ force on the box and move it 5 meters. The force is parallel to the floor, such that $$\theta=0$$. Calculate the work done on the box. Using the equation for work, we calculate that $$Work = 120 N \times 5 m\times cos(0)$$ $$Work = 600 J$$ Now consider these two scenarios: 1. You push the box in 50 seconds 2. You push the box in 100 seconds Which scenario took more power? Let us do the calculations for both scenarios: $$Power_1 = \frac{600 J}{50}$$ $$Power_1 = 12 W$$ $$Power_2 = \frac{600 J}{100}$$ $$Power_2 = 6 W$$ By comparing the two values of power, we see that scenario 1 took more power. Since power is a rate, the faster the work is done, the higher the rate and therefore the larger the power.

### Subject:Algebra

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Question:

How do you graph a function with absolute values? For example: $$y=|2x+3|$$

Inactive
Smiti N.

First, recall this property of absolute values: $$|a|=6$$ has two solutions: $$a=6$$ and $$a=-6$$ The first step when graphing absolute value functions is to split the function into two parts using that property. The function $$y=|2x+3|$$ can be split into: $$y=2x+3$$ (Case 1) and $$-y=2x+3$$ (Case 2) For Case 2, let us divide both sides of the equation by -1 to get $$y=2x+3$$ Now we have two equations: $$y=2x+3$$ and $$y=2x+3$$ Before we graph these, remember that absolute values have to be positive. Since our original function said that $$y$$ equals the absolute value of something, we know that $$y$$ must be positive. Now we can graph just the parts of $$y=2x+3$$ and $$y=2x+3$$ that are are positive (in quadrants I and II). The graph should look like a "V" and the tip of the V should be at $$(\frac{-3}{2}, 0)$$

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