Enable contrast version

# Tutor profile: Kyle H.

Inactive
Kyle H.
Math and Physics Tutor for 4 Years
Tutor Satisfaction Guarantee

## Questions

### Subject:Calculus

TutorMe
Question:

What is the arc length of the curve $$f(x) = 2x^{3/2}$$ on the interval $$0 \le x \le 1$$ ?

Inactive
Kyle H.

The arc length formula is: $$L = \int_a ^b \sqrt{1 + [f'(x)]^2} dx$$ First off, let's find [f'(x)]^2. We'll first need to take the derivative of $$f(x)$$ then square it: -- $$f(x) = 2x^{3/2}$$ Taking the derivative of $$f(x)$$ is as easy as using the power rule: -- $$f'(x) = 2\cdot (3/2) x^{3/2 -1}$$ -- $$f'(x) = 3x^{1/2}$$ Now let's square it: -- $$[f'(x)]^2 = (3x^{1/2})(3x^{1/2})$$ -- $$[f'(x)]^2 = (9x)$$ Now, let's plug everything into the arc length formula and run the integration. We're on the interval $$0 \le x \le 1$$, so that means that $$a$$ and $$b$$ are $$0$$ and $$1$$ respectively: -- $$L = \int_a ^b \sqrt{1 + [f'(x)]^2} dx$$ -- $$L = \int_0 ^1 \sqrt{1 + [f'(x)]^2} dx$$ Next let's insert $$[f'(x)]^2 = (9x)$$: -- $$L = \int_0 ^1 \sqrt{1 + 9x} dx$$ Now let's replace the square root with an equivalent 1/2 power: -- $$L = \int_0 ^1 (1 + 9x)^{1/2} dx$$ It looks like u-substitution will be the easiest way to integrate this function: -- $$u = 1+9x$$ --$$du = 9dx$$ --$$(1/9)du = dx$$ With $$u = 1+9x$$ we can plug in our bounds $$x=0, x=1$$ and solve for new bounds in terms of $$u$$: $$For x=0, u = 1$$ $$For x=1, u = 10$$ Now we have: -- $$L = \int_1 ^{10} (u)^{1/2} (1/9)du$$ -- $$L = (1/9) \int_1 ^{10} (u)^{1/2} du$$ This can be easily integrated using the reverse power rule: -- $$L = (1/9) (2/3)u^{3/2} |_1 ^{10}$$ Now let's just plug in the bounds and finish it off: -- $$L = (2/27) [ (10)^{3/2} - (1)^{3/2} ]$$ -- $$L = (2/27) [ 500 - 1 ]$$ -- $$L = (998/27)$$

### Subject:ACT

TutorMe
Question:

When $$x = 3$$ and $$y = 5$$, by how much does the value of $$3x^2 – 2y$$ exceed the value of $$2x^2 – 3y$$ ? A. 4 B. 14 C. 16 D. 20 E. 50

Inactive
Kyle H.

There are mainly two ways to do this problem. In the first method you you would plug $$x = 3$$ and $$y = 5$$ into $$3x^2 – 2y$$ and $$2x^2 – 3y$$ then take the difference between the two: -- $$3(3)^2 - 2(5) = 17$$ -- $$2(3)^2 - 3(5) = 3$$ -- $$17 - 3 = 14$$ For me, this first method is slightly slower. On a timed test like the ACT, you want to maximize your time, and the test writers almost always have an intended method of solution which is quick and easy. A slightly more elegant method is to first take the difference between $$3x^2 – 2y$$ and $$2x^2 – 3y$$, then plug in $$x = 3$$ and $$y = 5$$. This is an exercise in combining like terms: -- $$3x^2 – 2y - (2x^2 – 3y) = x^2 +y$$ (Notice how combining the like terms is simple and easy to do, and produces a clean and easily workable result. Often times the test writers intended for this to be the case) --$$(3)^2 + (5) = 14$$

### Subject:Physics

TutorMe
Question:

An electron traveling due North with constant speed $$v_e$$, passes through an electric field of magnitude $$E$$ pointing due East and a magnetic field with no deflection. What a) direction must the magnetic field be oriented and b) how strong is it relative to the electric field?

Inactive
Kyle H.

WARNING: This answer is much easier done than said. I can demonstrate this via video chat/in person much more easily than writing it all out. If your head sort of hurts after reading this, I'm sorry. An electron going through an electric field will experience a Coulomb force: $$\vec F_E = q\vec E$$. An electron having a negative charge will be deflected opposite the direction of the electric field. So if only the electric field were present, the electron would be deflected to the West with a force of magnitude $$|F_E| = qE$$ A magnetic field will act on a moving charge according to the Lorentz force: $$\vec F_B = q \vec{v} \times \vec B$$. In order for the electron to experience no deflection overall, this Lorentz force must have the same magnitude as the Coulomb force, but act in the opposite direction. The right hand rule can be used to determine the direction of the magnetic field. To find the direction the Lorentz force acts, you first point the fingers of your right hand the direction of travel of the electron, then curl your fingers into the direction the magnetic field points. Because we're dealing with a negative charge, you then need to turn your hand so your thumb is pointing opposite the direction it was. After all of that , your thumb will be pointing the direction the force will act. So, working backwards, we can start with our thumb pointed due East (the direction we know that the force needs to act) and our fingers curled. Then we'll flip our hand over so our thumb is pointed in the opposite direction. Next, we extend our fingers out to point North (the direction of travel of the electron). While uncurling our fingers they point upward, so that is the direction the B field must point for the Lorentz force to act due east on this electron! To find the strength of the magnetic field relative to the electric field, we'll set the Coulomb force and the Lorentz force equal: -- $$F_E = F_B$$ -- $$qE = qvB$$ (Note: The magnitude of a cross product can be expressed as the magnitudes of the vectors involved multiplied by the sine of the angle between them. So: $$|q\vec v \times \vec B| = qvBsin\theta$$ The angle between due north and straight up is $$90^\circ$$. $$sin90^\circ = 1$$, so $$qvBsin\theta \rightarrow qvBsin 90 ^\circ = qvB$$) -- $$\frac{E}{v} = B$$ To sum up: a) The magnetic field must point straight up b) The magnitude of the magnetic field must be $$\frac{E}{v}$$

## Contact tutor

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

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.
BEST IN CLASS SINCE 2015
TutorMe homepage