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Guruprasad R.
Graduate Student at Caltech
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MATLAB
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Question:

The perfect gas law relates the pressure p ($$\frac{N}{m^2}$$) , absolute temperature T (K), mass n (moles) and volume V ($$m^3$$) of a gas. It states that $$ pV = nRT $$ where the constant R is the gas constant. The value of R for air is 286.7 Nm/kgK. Suppose air is contained in a chamber at room temperature ($$ 0^oC $$ = 273K). A. Create a function that computes the gas pressure p in $$\frac{N}{m^2}$$ from input arguments V and n. Ensure your function accepts V as a range of values.

Guruprasad R.
Answer:

The function in MATLAB to solve this question is presented below with comments to help with understanding. %Task (Function to output values of P) function [P] = task3_evaluatePressure (V,m) %P is the output variable, and (V,m) are the input %variables R = 286.7; %Given T = 293; %Given P = m*R*T./V; %./ is the element wise division element as V is a range of values {follows from formula} end

Linear Algebra
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Question:

$$ W= {(x,y) \in R^2 | xy \geq 0 } $$ Which of the following statements is true: a) W is a subspace b) W is not a subspace as it isn't closed under vector addition c) W is not a subspace as it isn't closed under scalar multiplication d) W is not a subspace as it doesn't contain the zero vector

Guruprasad R.
Answer:

To determine if W is a subspace or not, there are 3 major criteria one must verify: i) W should have the zero vector $$ \underline{Proof}: $$ Let element (0,0) belong to W. Does it abide by the conditions imposed? Here, the condition is (0,0) $$ \in R^2 $$ and $$ 0*0 \geq 0 $$. In this case, both the conditions are satisfied, which implies it does contain the zero vector. ii) W should be closed under vector addition $$ \underline{Proof}: $$ Let {$$ (x_1, y_1), (x_2, y_2) $$} belong to the W space. This implies that both the conditions imposed must be met by the elements {$$ (x_1, y_1), (x_2, y_2) $$}. In other words: $$ (x_1, y_1) \in R^2; (x_2, y_2) \in R^2 $$ and $$ x_1*y_1 \geq 0; x_2*y_2 \geq 0$$ Now, to verify the vector addition of the 2 elements lie within the W-space. Vector addition of the 2 elements ==> $$ (x_1 + x_2, y_1 + y_2) $$ To show they belong to the W-space, we need to prove that $$ (x_1+x_2)*(y_1+y_2) \geq 0 $$ $$ \begin{equation} = (x_1 + x_2)*(y_1 + y_2) \\ = (x_1*y_1 + x_2*y_2 + x_1*y_2 + x_2*y_1) \\ \end{equation} $$ We know that the first 2 terms $$ (x_1*y_1 + x_2*y_2 ) \geq 0 $$, but the last 2 terms could be less than 0 - resulting in the expression evaluating to a negative value. As there are cases when the vector addition isn't closed, W is not a subspace. iii) W should be closed under scalar multiplication $$ \underline{Proof}: $$ Let {$$ (x_1, y_1) $$} belong to the W space. This implies that both the conditions imposed must be met by the element {$$ (x_1, y_1) $$}. In other words: $$ (x_1, y_1) \in R^2 $$ and $$ x_1*y_1 \geq 0 $$ Now, to verify if the scalar multiplication of the element lies within the W-space. Let 'c' be a scalar and it's product with the element is: $$ (c*x_1, c*y_1) $$ To show this element belongs to the W -space, we need to show that: $$ \begin{equation} (c*x_1) * (c*y_1) \geq 0 \\ = c*c * x_1*y_1 \\ = c^2 *(x_1 * y_1) \end{equation} $$ For any scalar $$ c \in R $$, $$ c^2 \geq 0 $$ AND $$ x_1*y_1 \geq 0 $$. Therefore the product $$ c^2 *(x_1 * y_1) \geq 0 $$ This shows that W is closed under scalar multiplication. Given this analysis: we can see that the only true statement is ; (b) - W is not a subspace as it isn't closed under vector addition

Calculus
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Question:

Q. Solve this expression: $$\int_{5}^{5} \frac{(x^2-e^x)}{sin^2(x)} dx$$ and explain your reasoning as well.

Guruprasad R.
Answer:

Before, we jump to evaluating the expression by applying relevant formulas, the limits of the integral should be noticed. The integral of an expression (here - $$\frac{(x^2-e^x)}{sin^2(x)} $$) implies the evaluation of the area under the curve between the limits of integration. As in the question, if the limits are from 5 to 5, whatever be the function (curve), the area under the curve will be 0. Therefore, without having to evaluate the complex expression, we arrive at the answer - 0.

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