Enable contrast version

Tutor profile: Spencer P.

Inactive
Spencer P.
Computer Science Student at UCLA
Tutor Satisfaction Guarantee

Questions

Subject: Calculus

TutorMe
Question:

$$y = x^3 + \sin(2x)$$. Find $$\frac{dy}{dx}$$.

Inactive
Spencer P.
Answer:

We can find $$\frac{dy}{dx}$$ by deriving each term of the right hand side of the given equation separately. That is, $$\frac{dy}{dx} = \frac{d}{dx}[x^3 + \sin(2x)] = \frac{d}{dx}[x^3] + \frac{d}{dx}[\sin(2x)]$$. Using the power rule, we know $$\frac{d}{dx}[x^3] = 3x^2$$. To find $$\frac{d}{dx}[\sin(2x)] $$ we use the chain rule. Let's define a new variable $$u$$, where $$u = 2x$$. Then we can rewrite $$\frac{d}{dx}[\sin(2x)] $$ as $$\frac{d}{dx}[\sin(u)] $$. Now it's easy to find $$\frac{d}{du}[\sin(u)] $$, which is defined to be $$\cos(u)$$. However, we want the derivative with respect to $$x$$, not the derivative with respect to $$u$$. This is where we use the chain rule, which tells us $$\frac{d}{dx}[\sin(u)] = \frac{d}{du}[\sin(u)] \cdot \frac{du}{dx} = \cos(u)\cdot \frac{du}{dx}$$. What's $$\frac{du}{dx}$$? Since $$u = 2x$$, it's just $$\frac{d}{dx}[2x]$$ = 2. So $$\frac{d}{dx}[\sin(2x)] = \cos(u) \cdot 2 = 2\cos(2x)$$. So our final answer is $$\frac{dy}{dx} = 3x^2 + 2\cos(2x)$$.

Subject: Physics

TutorMe
Question:

Using Gauss's Law, find the magnitude of the electric field of a point charge $$q$$ as a function of the distance $$r$$ from the point charge.

Inactive
Spencer P.
Answer:

Let's call our surface $$S$$. Gauss's Law states that the electric flux through a surface is proportional to the charge enclosed inside the surface: $$\oint_S{\vec{E}}\bullet{\vec{dA}} = \frac{q_{encl}}{\epsilon_0}$$ However, we're trying to find the electric field at a point in space. To solve this problem we imagine a spherical surface of radius $$r$$ centered around the point charge. Then due to symmetry, Gauss's Law simplifies and allows us to find the electric field. Since $$\vec{E} \perp \vec{dA}$$ at all points on the surface, $$\oint_S{\vec{E}}\bullet{\vec{dA}} = E(r) \cdot A$$, where $$A$$ is the surface area of $$S$$. $$E(r)$$ is what we're solving for, and the surface area $$S$$ of our sphere is $$4\pi r^2$$. Our enclosed charge $$q_{encl}$$ is just $$q$$, since we're assuming the point charge is the only charge inside our surface. So now we have: $$E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0} $$ Isolating $$E(r)$$ gives us our solution, $$E(r) = \frac{q}{4\pi r^2\epsilon_0} $$.

Subject: Differential Equations

TutorMe
Question:

The following differential equation has distinct, real, characteristic roots. Find the general solution: 2y'' - y' - 6y = 0

Inactive
Spencer P.
Answer:

This is a second-order linear homogeneous differential equation, so there are two solutions of the form y($$t$$) = $$e^{\lambda t}$$, where the two values of $$\lambda$$ are given by solving the equation $$2\lambda^2 - \lambda - 6 = 0$$. This "characteristic polynomial" is a quadratic equation found by replacing different order derivatives of y in the equation given with corresponding powers of $$\lambda$$. We can solve the equation by factoring: $$2\lambda^2 - \lambda - 6 = 0$$ $$(2\lambda + 3)(\lambda - 2) = 0$$ $$2\lambda + 3 = 0$$ or $$\lambda - 2 = 0$$ $$\lambda = -\frac{3}{2}$$ or $$\lambda = 2$$ So two specific solutions that satisfy the given equation are y($$t$$) = $$e^{-\frac{3}{2} t}$$ and y($$t$$) = $$e^{2t}$$. Since these solutions are linearly independent, the general solution is all linear combinations of these two solutions: y($$t$$) = $$C_1e^{-\frac{3}{2} t} + C_2e^{2t}$$, where $$C_1$$ and $$C_2$$ are any two integers.

Contact tutor

Send a message explaining your
needs and Spencer will reply soon.
Contact Spencer

Request lesson

Ready now? Request a lesson.
Start Lesson

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 Lesson" 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.