# Tutor profile: Thomas W.

## Questions

### Subject: Linear Algebra

a.) Determine if the function below is a linear function. Show all supporting work: $( f(x)=2x+3 $) b.) How does your conclusion compare to your understanding of linear function from past math classes?

a.) In linear algebra, a linear function must satisfy both of the following conditions: $( 1. \space f(x+y)=f(x)+f(y) $) $(2. \space f(cx)=cf(x) $) Let us check these two conditions on the given function by substitution: Condition 1: $( f(x+y) \stackrel{?}{=} f(x)+f(y) $) $( 2(x+y)+3 \stackrel{?}{=} 2(x)+3 +2(y)+3$) $( 2x+2y+3 \neq 2x+2y+6$) Already we see that the first condition is not met, so the function $$f(x)$$ is not a linear function based on the definition. b.) This is different that analysis in previous math classes because same words "linear function" have been used to describe two different kinds of relationships. In high school mathematics and calculus, a linear function is anything with a constant rate of change, so $$f(x)$$ would have been classified as a linear function in those settings, but using our new, different linear algebra definition, $$f(x)$$ is not a linear function.

### Subject: Geometry

Find the volume of the pyramidal solid $$S$$ with height $$h$$ and an equilateral triangle base with side length $$a$$.

To calculate the volume of the solid, we must first recall the fomula for the volume of a pyramid with a general base $$B$$. $( V=\frac{1}{3}Bh$) From this formula, we only need to substitute the height of the solid (which is given: $$h$$) and the area of the base $$B$$. Furthermore, it tells us the base is an equilateral triangle with side length $$a$$. If we recall the area of a triangle is equal to $$\frac{1}{2}b_th_t$$ where $$b_t$$ and $$h_t$$ are the base and height of the triangle respectively. The base of the triangle will simply be $$a$$, but to find the height of the triangle, we will need to find the altitude of the triangle. If we draw and altitude on an equilateral triangle, we find that it bisects the triangle, creating two 30-60-90 special right triangles. As the height of the triangle is the longer leg of the special right triangle, we can apply either the special right triangle relationships or trigonometry to find $$h_t = \frac{a\sqrt{3}}{2}$$. Now we simply substitute into our area formula: $( B=\frac{1}{2}b_th_t=\frac{1}{2}(a)(\frac{a\sqrt{3}}{2}) =\frac{a^2\sqrt{3}}{4} $) Now that we have the area of the base, we can simply substitute into the volume formula: $( V= \frac{1}{3}Bh=\frac{1}{3}(\frac{a^2\sqrt{3}}{4})(h) = \frac{a^2h\sqrt{3}}{12} $)

### Subject: Calculus

How does the derivative of a function $$ \frac{d}{dx}f(x) $$ relate to the gradient of a function of 2 variables $$ \nabla g(x, y) $$

The gradient of a 2 variable function is a more generalized rate of change than the 1 variable case. Specifically, the gradient tells the rate of change as you look in both the x and y directions. In contrast the derivative of the 1 variable function only tells the rate of change along the x axis - as that is the only axis on which the function $$ f $$ can vary. This idea of the gradient being a more generalized derivative is even more obvious when you look at the geometric interpretations of both of these "derivatives." In the 2 variable case, the derivative (gradient) is represented by a two dimensional vector where as in the 1 dimensional case, it is only represented by a single number. This is because the gradient vector has the rates of change with respect to both of the variables in the function ( x and y), rather than just for the single variable. Geometrically, students are initially taught that the geometric interpretation of a derivative is the slope of the tangent line to the function at a specific point. This analogy can be extended when the number of variables is increased as the magnitude of the gradient represents the largest rate of change of the function at a given point and the gradient vector itself tells the specific direction in which that largest rate of change happens. Altogether, the gradient provides the information to create all tangent lines the function of 2 variables at a given point.