# Tutor profile: Milan S.

## Questions

### Subject: SAT

$$ x^2 + y^2 = 153 $$ $$ y = -4x $$ If (x,y) is a solution to this system, what is $$x-1$$?

In SAT math questions like this, always remember what the question is asking! We want $$x-1$$. Let's go ahead and solve the system. If we substitute $$ y = -4x $$ into the first equation we will get: $$ x^2 + (-4x)^2 = 153 $$ $$ x^2 + 16x^2 = 153 $$ $$ 17x^2 = 153 $$ $$ x^2 = 9 $$ $$ x = -3 $$ or $$x = 3$$ $$ x -1 = -4 $$ or $$ x - 1 = 2$$ Be careful, only one of these solutions will be in the answer choices and you must choose the right one! Alternatively, -3 or 3 will definitely be in the answer choices in case you forget that we want $$x-1$$ not $$x$$!

### Subject: Computer Science (General)

With k sorted linked lists, describe an efficient algorithm that returns the sorted list of all nodes in all k linked lists.

The naïve approach is to use the mergesort algorithm which has a runtime of O(nlog(n)), where n is the total number of nodes present. A better approach would be to construct a min-heap of all of the heads of the k linked lists. We will pop off the minimum and add it to our output. We will run this again to see which is the next minimum, and we will continue popping off the minimum from all of the lists until all lists our empty. What runtime does this approach have? O(nlog(k)). Why? Because we are still going through all of the n nodes, but at each pass we are reordering our min-heap. This means we are performing a log(k) operation each time, hence O(nlog(k)).

### Subject: Calculus

If given a graph that models velocity over time (where the x-axis is time and the y-axis is velocity), define what the integral of this function really means. Explain in as much detail as possible.

If we take the integral of a function v(t), where v(t) is the velocity of a particle, we will be adding up the area under the velocity curve. The area under the velocity curve will give us the position of our particle. Consider the function: v(t) = 5t. This means that at t=0, our particle is at rest (zero velocity). If we wish to integrate this function, we will get the position function x(t). x(t) = $$ \int \mathrm{5t} \mathrm{d}t $$ x(t) = $$ \frac{5}{2}t^2 + C $$ The constant C in this case is simply based on where our particle starts. Let's assume it starts at the origin. Then: x(t) = $$ \frac{5}{2}t^2 $$ With this function, we can tell what the position of our particle is over time. At t=1, the particle has moved a distance of 2.5 m, then at t=2 it has moved 10 m, then at t=3 it has moved 22.5 m, and so on. The point of this exercise is to understand what an integral really means. Is it just an anti-derivative? It really is the sum of the area under a curve, and in this case we have shown a physical interpretation of this. Try this again with the concept of derivatives and see if you can make the connection between position and velocity!

## Contact tutor

needs and Milan will reply soon.