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# Tutor profile: Chris M.

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Chris M.
Mathematics Tutor for 5 years
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## Questions

### Subject:Linear Algebra

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Question:

Compute the following: $$det \left( \left[ \begin{array}{ccc} 3 & 4 & 5 \\ 1 & 3 & 7 \\ 3 & 8 & 3 \end{array} \right] \left[ \begin{array}{ccr} 2 & 1 & 4 \\ -2 & 5 & 5 \\ 1 & -1 & -1 \end{array} \right] \right)$$

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Chris M.

There are two solution routes to take. This comes from the properties of determinants of square matrices. Solution Method 1: $$\left[ \begin{array}{ccr} 3 & 4 & 5 \\ 1 & 3 & 7 \\ 3 & 8 & 3 \end{array} \right] \left[ \begin{array}{ccr} 2 & 1 & 4 \\ -2 & 5 & 5 \\ 1 & -1 & -1 \end{array} \right] = \left[ \begin{array}{ccc} 3(2) + 4(-2) + 5(1) & 3(1) + 4(5) + 5(-1) & 3(4) + 4(5) + 5(-1) \\ 1(2) + 3(-2) + 7(1) & 1(1) + 3(5) + 7(-1) & 1(4) + 3(5) + 7(-1) \\ 3(2) + 8(-2) + 3(1) & 3(1) + 8(5) + 3(-1) & 3(4) + 8(5) + 3(-1) \end{array} \right]$$ $$= \left[ \begin{array}{ccc} 6 + -8 + 5 & 3 + 20 + -5 & 12 + 20 + -5 \\ 2 + -6 + 7 & 1 + 15 + -7 & 4 + 15 + -7 \\ 6 + -16 + 3 & 3 + 40 + -3 & 12 + 40 + -3 \end{array} \right]$$ $$= \left[ \begin{array}{ccc} 3 & 18 & 27 \\ 3 & 9 & 12 \\ -7 & 40 & 49 \end{array} \right]$$ Now take the determinant: $$det \left( \left[ \begin{array}{ccc} 3 & 18 & 27 \\ 3 & 9 & 12 \\ -7 & 40 & 49 \end{array} \right] \right)$$ $$= 3 \left[ 9(49) - 40(12) ] - 18 [ 3(49) - 12(-7) ] + 27 [ 3(40) - 9(-7) \right]$$ $$= 3(441 - 480) - 18(147 + 84) + 27(120 + 63)$$ $$= 3(-39) - 18(231) + 27(183)$$ $$= 666$$ Solution Method 2: Multiplying two 3x3 matrices before computing the determinant is often times very time consuming. Since both matrices are the same size and square, the following property can be used: $$det(AB) = det(A)det(B)$$ Then, the problem becomes easier to compute. $$det(A) = det \left( \left[ \begin{array}{ccc} 3 & 4 & 5 \\ 1 & 3 & 7 \\ 3 & 8 & 3 \end{array} \right] \right)$$ $$det(A) = 3[3(3) - 7(8)] - 4[1(3) - 7(3)] + 5[1(8) - 3(3)]$$ $$det(A) = 3(9 - 56) - 4(3 - 21) + 5(8 - 9)$$ $$det(A) = 3(-47) - 4(-18) + 5(-1)$$ $$det(A) = -74$$ $$det(B) = det \left( \left[ \begin{array}{ccr} 2 & 1 & 4 \\ -2 & 5 & 5 \\ 1 & -1 & -1 \end{array} \right] \right)$$ $$det(B) = 2[ 5(-1) - 5(-1) ] - 1 [ -2(-1) - 5(1) ] + 4[ -2(-1) - 5(1) ]$$ $$det(B) = 2(-5 + 5) - 1(2 - 5) + 4(2 - 5)$$ $$det(B) = 2(0) - 1(-3) + 4(-3)$$ $$det(B) = -9$$ Then, the final solution $$det(AB) = (-74)(-9) = 666$$

### Subject:Calculus

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Question:

Why does the evaluation of a derivative give the slope of the tangent line to it's point?

Inactive
Chris M.

This is a relatively simple concept that is extremely important to understand for all future work in calculus. The slope of a line is defined as $$\frac{\Delta y}{\Delta x}$$ I can make any changes to how large my $$\Delta y$$ is, and since we are using the line example, the size of $$\Delta x$$ will also change with it. With both values changing accordingly, the slope of the line will remain the same! Now, rather than making $$\Delta y$$ larger, we can shrink it's size until we get something indiscernible in size. $$\Delta y \Rightarrow dy$$ Again, since we have no change in the value of the slope of a line, decreasing the value of $$\Delta y$$ will also decrease the value of $$\Delta x$$. Thus, $$\Delta x \Rightarrow dx$$. Then, the slope of the line can be rewritten as follows! $$Slope = \frac{\Delta y}{\Delta x} = \frac{dy}{dx}$$ Now, we know that the derivative of a function y with respect to x is $$\frac{dy}{dx}$$, and so evaluating a derivative at some point x gives a slope value of a line containing this point.

### Subject:Physics

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Question:

In a roller coaster loop, do you "feel" heavier at the top or the bottom of the loop?

Inactive
Chris M.

Your sense of weight comes from objects pushing back on you, known as the normal force. When cresting a hill in a car at high enough speeds, you feel a momentary loss in weight. The same effect occurs in roller coasters when reaching peaks. However, in a roller coaster loop, the forces at play become more complicated. The question asks at which point, the top or bottom, do you "feel" heavier. In other words, at which point of the coaster loop is the Normal Force greatest? At every point along the ride, your total mass combined with gravity create the Weight Force that is continuously pointed downward. As you progress through the loop and enter circular motion, the Net Force acting on you must point towards the center of the loop to continue with the motion. This net force is called the Centripetal Force, and it is commonly misunderstood. At the top of the loop, the centripetal force points downward in the same direction as gravity. It is important to again understand that the centripetal force IS the net force, and that the gravitational force (your weight) is a "contribution" to this net force. Gravity alone cannot provide enough force to attribute to the total centripetal force, so the rest is supplied by your chair pushing back on you! This is the normal force, and it is your "Feeling" of weight. At the bottom of the loop, gravity is again pointed downwards, but this time the centripetal force is pointed upwards. The normal force here will be greater than the normal force from the top of the loop because of this difference in orientation. The following equation can be considered: Fcentripetal = Fnormal + Fgravity So, to answer the question, at the Bottom of the Coaster loop, you will feel heavier than when you are at the top of the loop! Just remember that when solving circular motion problems, that the depicted images of Centripetal Force pointed towards a center of rotation should be seen as the Net Force acting on the object in motion.

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