# Tutor profile: Kurt B.

## Questions

### Subject: Calculus

This is a calculus 3 concept: Evaluate the partial derivatives of the following function-- f(x,y) = 6e^(2x+y) + 2y*ln(3y-x)

-Lets find the "x" partial derivative: fx(x,y) = 12e^(2x+y) + (-2y/(3y-x)) This is found by treating any "y's" as coefficients. Almost think of them as placeholders for a number. -Repeat for "y" partial: fy(x,y) = 6e^(2x+y) + {(2*ln(3y-x) + (2y*(1/(3y-x)*(3))} We needed to use chain rule and product rule this time around, make sure to keep track of those variables. Cleaned up: fy(x,y) = {6e^(2x+y) + (6y/(3y-x))}

### Subject: C++ Programming

I'm sure you all would love to see some fancy recursion but lets keep it simple :) Q: Print the statement "Hello World" While we could do this in a very simple "cout << ... ;" command, lets show off arrays and print a list of characters stored within one. Comments are denoted by beginning with "//" as is used in c++.

# include <iostream> using namespace std; int main() { //Declaring a character array to hold letters and characters that spell "Hello World". char array[11]{ 'H','e','l','l','o',' ','W','o','r','l','d' }; //Using a for loop and a stepper variable, we print the letter held in the array value of i //i increases until i = 11 and terminates the loop. for (int i = 0; i < 11; i++) { cout << array[i]; } //creates new line to print the system statement "pause". cout << endl; //follows good coding practice and calls a pause command before terminating the program. system("pause"); //terminates code. return 0; }

### Subject: Physics

A camera lens used for taking close-up photographs has a focal length of 25.0 mm. The farthest it can be placed from the film is 31.5 mm. What is the closest object (in mm) that can be photographed?

- The basic formula for any lens (convex or concave) is as follows: 1/focal length = (1/distance image) + (1/distance object) - In the problem above, our focal length is stated to be 25.0 mm. The distance from the lens to the image created is 31.5 mm. If we are to solve for the distance from the lens to the object, let's rearrange our equation to solve for the unknown focal length = f, distance object = dO, distance image = dI {1/f = (1/dI) + (1/dO) } ==> { 1/f - 1/dI = 1/dO} ==> dO = (1/f-1/dI)^-1 dO = [ 1/(25 mm) - 1/(31.5 mm) ] ^-1 dO = 121.2 mm