Tutor profile: Tyler R.
Subject: Java Programming
Implement the following relationship if possible: The class Red has variable rValue. The class Green has variable gValue. The class Blue has variable bValue. Finally, the class Yellow extends and inherits from both classes Red and Green.
This relationship is not possible. Java does not support multiple inheritance, meaning the class Yellow could not extend both class Red and Green at the same time. However, if Red and Green were interfaces, the class Yellow could implement both simultaneously. A more optimal solution to this problem would be to either have Red, Green, and Blue as interfaces or have all colors inherit from a single parent class or interface that contains all three variables rValue, gValue, and bValue. Each class of color could specify the values of these inherited variables within their respective constructors.
Subject: Computer Science (General)
What is the difference between the data structures stack and queue?
A stack operates in a LIFO fashion (Last In First Out). Think of a stack as a literal stack of objects. For example, you stack 10 computer science textbooks on a table. The first textbook you placed would be the one at the bottom (the least accessible). However, the last textbook placed would be the one at the top (the most accessible). If you wanted to get to the seventh book you placed down (the third from the top), you would have to remove the first 2 textbooks first. Similarly, in the stack data structure, objects can be placed into a stack easily and the last object placed can be immediately removed from the stack. However, removal of the first object inserted would require the removal of all of the objects inserted afterward. A queue works the opposite, known as FIFO (First In First Out). Imagine a line of customers at a bank. The first customer to arrive would be the first one tended to, and a line would build behind them. If you came in at the end, you would be last in the line and thus the last to be tended to. Just like this hypothetical line of people, a queue allows for objects to be inserted at the end and retrieved from the front. So, after inserting 10 items, removal of an item from the queue would give you the first item you inserted and the second item inserted would now become the head of the queue.
A 10ft ladder is sliding down a wall. The base of the ladder is 6ft from the wall. If the top of the ladder is sliding down the wall at a rate of 2ft/s, how fast is the base of the ladder moving away from the wall?
For most word problems in calculus, the first step is to visualize it. The best way to illustrate this problem would be drawing a right triangle. The base x represents the distance between the ladder's base and the wall, the height y represents the height at which the top of the ladder touches the wall, and the hypotenuse s represents the ladder itself. We know that s = 10ft and x = 6ft. By using Pythagoras's theorem (s^2 = x^2 + y^2), we can find y = sqrt(s^2 - x^2) = sqrt((10ft)^2 - (6ft)^2) = 8ft. In addition, we have the knowledge that the top of the ladder is sliding down the wall at rate dy = -2ft/s (the rate is negative because y is established to be the height at which the top of the ladder touches the wall, so when sliding down this y would grow smaller, thus having a negative rate). The fact that the ladder is moving and rates are involved should alert us to the notion that a derivation must be made to find the rate dx. Reverting back to Pythagoras's theorem, we can find that implicitly deriving the equation produces a new equation with all of our necessary variables. So, by implicitly deriving Pythagoras's theorem, we get that 2sds = 2xdx + 2ydy. Since the ladder's length (s) is not changing, ds must be 0. Therefore, 0 = 2xdx + 2ydy or, since we're solving for dx, dx = 2ydy/-2x. Finally, dx = -ydy/x. Plugging in our variables yields dx = -(8ft)(-2ft/s)/(6ft) = 2.667ft/s. Therefore, the base of the ladder is moving away from the wall at a rate of 2.667ft/s.
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