TutorMe homepage
Subjects
PRICING
COURSES
SIGN IN
Start Free Trial
Guruprasad R.
Graduate Student at Caltech
Tutor Satisfaction Guarantee
MATLAB
TutorMe
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
TutorMe
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 $$(x_1+x_2)*(y_1+y_2) \geq 0$$ $$$$= (x_1 + x_2)*(y_1 + y_2) \\ = (x_1*y_1 + x_2*y_2 + x_1*y_2 + x_2*y_1) \\$$$$ We know that the first 2 terms $$(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)$$ To show this element belongs to the W -space, we need to show that: $$$$(c*x_1) * (c*y_1) \geq 0 \\ = c*c * x_1*y_1 \\ = c^2 *(x_1 * y_1)$$$$ For any scalar $$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
TutorMe
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.

Send a message explaining your
needs and Guruprasad will reply soon.
Contact Guruprasad
Ready now? Request a lesson.
Start Session
FAQs
What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Session" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
Made in California
© 2019 TutorMe.com, Inc.