# Tutor profile: Maiya L.

## Questions

### Subject: Calculus

Solve the initial-value problem y'' + 10y' + 16y = 0 y(0) = 2 and y'(0) = -10

Let y = e^(rt) and substitute into the given equation to get (e^(rt))'' + 10(e^(rt))' + 16e^(rt) = 0 - - equation 1. The derivative of e^(rt) is r*e^(rt) and the second derivative of e^(rt) is r^2*e^(rt). Therefore, because of the derivatives and dividing everything by e^(rt) which will never be 0, equation 1 simplifies into r^2 + 10r + 16 = 0 which is also (r+2)(r+8) = 0. So r = -2 and -8. Recall, the general solution to a differential equation is y(t) = c1*e^(rt) + c2*e^(rt), where c1 and c2 are coefficients. The general solution to this problem is y(t) = c1*e^(-2t) + c2*e^(-8t). So, y'(t) = -2*c1*e^(-2t) - 8*c2*e^(-8t). If we plug in the initial conditions, we get 2 = c1*e^(0) + c2*e^(0) which simplifies into 2 = c1 + c2 or c1 = 2 - c2 and -10 = -2*c1*e^(0) -8*c2*e^(0) which simplifies into -10 = -2*c1 - 8*c2. Plugging c1 = 2 - c2 into -10 = -2*c1 - 8*c2, we get -10 = -2*(2 - c2) - 8*c2. Solve for c2 c2 = 1, so c1 = 2 - 1 = 1. Therefore the solution to the initial value problem is y(t) = e^(-2t) + e^(-8t)

### Subject: Calculus

Determine the critical points of the following function: F(x) = -2x^5 + 5x^4 + 80x^3 + 1

The first step is to take the first derivative of F(x). Recall that when you take the derivative of a function, you want the order of magnitude for each term to decrease by 1. The derivative of F(x) is F'(x) = -2*5x^(5-1) + 5*4x^(4-1) + 80*3x^(3-1) + 1*0 which simplifies into F'(x) = -10x^4 + 20x^3+ 240x^2. Note: the ' in the F'(x) means it is the derivative of F(x). You can factor -10x^2 out and get F'(x) = -10x^2 * (x^2 - 2x - 24). This can be simplified into F'(x) = -10x^2 * (x + 4)(x - 6). When working on problems involving the maximum, minimum, critical points, etc. always try to factor and simplify as much as possible if you are able to. Remember that critical points are the points where the derivative is zero and/or does not exist. Because our derivative is a polynomial and it exists everywhere, we need to find where the derivative equals zero in order to find the critical points. In our case it is easy to identify which x values make F'(x) = 0. F'(x) = 0 when x = 0, x = -4, and x = 6, which are the critical points.

### Subject: Algebra

Sam is an only child. The sum of his father's age, twice his mother’s age, and three times his age is 95. The sum of two times his fathers age, four times his mother's age, and six times his age is 170. The sum of five times his father's age, ten times his mother's age and fifteen times his age equals 450. How many possible solutions of how old Sam and his parents are there?

Let Sam's age be called S, his father's age be called F, and his mothers age be called M. First, write an equation for each line of clues given. The first clue was that Sam's father's age + 2x his mother's age + 3x Sam's age = 95 so we can write that in an equation using symbols. F + 2M + 3S = 95 - - Equation 1 The second clue is that 2x his father's age + 4x his mother's age + 6x Sam's age = 170. 2F + 4M + 6S = 190 - -Equation 2 The third clue is that 5x his father's age + 10x his mother's age + 15x Sam's age = 450. 5F + 10M + 15S = 450 - - Equation 3 Notice how in equation 1 the coefficients on the left side are a factor of 2 different from the coefficients on the left side of equation 2. This means that if we multiply equation 1 by 2, then the left side of equation 1 should equal the left side of equation 2 and the right side of equation 1 should equal the right side of equation 2. If we multiply equation 1 by 2, we get 2F + 4M + 6S = 190 Now equation 1 matches equation 2 exactly. Next, we need to see if equation 1 or 2 can be manipulated to equal equation 3. Notice how in equation 1 the coefficients on the left side are a factor of 5 different from the coefficients on the left side of equation 3. This means that if we multiply equation 1 by 5, then the left side of equation 1 should equal the left side of equation 3 and the right side of equation 1 should equal the right side of equation 3. If we multiply equation 1 by 5, we get 5F + 10M + 15S = 475 However, 475 does not equal 450. This implies that there are no ages that you can plug into F, M, or S to satisfy both equations 1 and 3. Therefore there are no solutions to this problem.

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