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Degree: Physics with Theoretical Physics (Masters)  Imperial College London University
About Me:
My name is Ifan, I'm originally from North WalesÂ and I have recently graduated from Imperial College London with a 1st class honours masters degree in Physics with Theoretical Physics. I amÂ dedicated toÂ and have a trueÂ passion for the fields ofÂ science and mathematics; I am hoping to studyÂ forÂ a PhD in physicsÂ next year.Â My tutoring style is enthusiastic, creative and proactive, and I'm certain that anyÂ student of mineÂ will eventually shareÂ myÂ enthusiasm!Â I truly believe with a strong desire to learn and improve oneself, alongside the right guidance, any student can achieve great results.
My Experience
I am a veryÂ patientÂ andÂ friendlyÂ tutor; personally, I love teaching and helping students exceed their expectations.Â I have twoÂ yearsÂ of prior tutoring experience tutoringÂ two differentÂ students of varying aptitude in ALevel mathematics and physics. I also currently tutor my younger cousin in GCSE mathematicsÂ weekly. IÂ really enjoyed passing on my knowledge and getting to know my students, and alsoÂ achievedÂ great results.Â
The Sessions
My tuition is tailored to suit the student, accounting for their longterm goals (their desired career/profession) and shortterm goals (acing the exams!). I then break the work down into manageable steps and focus on the specific areas of the subject that the student requires extra help with, working alongside them until they are confident enough in their understanding to complete the task alone.Â With physics/maths, fundamental understanding is key to achieving great results in practice, so I ensure students have a thorough conceptual understanding of the topic before focussing on exam techniques. I believe this improves results, as opposed to simply memorising certain methodology.Â I also try to make learning fun and to have a bit of rapport with my students, as I believe this facilitates improved interaction and promotes a more relaxed and constructive learning environment.Â
What next?
If you have any questions, send me a 'WebMail' or book a 'Meet the Tutor Session'Â (both accessible through this website). Remember to mention your subject andÂ what you're struggling with.
Talk soon!
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(a) A frequency of 440 Hz means that a single trough and peak of the soundÂ wave oscillates 440 times per second or equally that 440 sound waves are produced by the keyboard per second (This is because the SI unit of a HzÂ is inverse seconds s^1)
(b) We first need the equation that relates the two given quantities we have (wave velocity and frequency) to the third unknown, desired quantity (wavelength). This is equation is a simple one, vÂ = f lambda. We want the quantity lambda, so we have to divide both sides of the equation by f to get v / f = lambda. We then plug in the numbers to find lambda, the wavelength, which is lambda = 340 m /s / 440 /sÂ = 340/440 m (as the /s units cancel out leaving just length which is the correct unit for a wavelength!)
see moreFirstly, we note that z being a complex number can be expressed in the formÂ z = a + bi. If we then take the complex conjugate of this expression, the real numbers remain the same (as they are their own c conjugates) but the c conjugate of i is i, therefore z* = a  bi. We then insert these expressions into the given equation, so
5 i (a + bi) Â + 3 (a  bi) + 16 = 8 i
We then expand the brackets and rearrange, remebering that i^{2}Â = 1, such that
5 i a  5 b + 3 a  3 b i + 16 = 8i.
We can then split the equation up into real and complex parts (i.e the terms that are not functions of i and are functions of i respectively)Â and treat each equation seperately, so we have
5 b + 3 a + 16 = 0, Â Â Â 5a  3b  8 = 0
This is a simple simultaneous equation to solve, and we this find that
16b = 104, 16a = 88
so z is thus given by z = 11/2 + 13/2 iÂ
QED
see moreSo we initially have the relationship y = ln(sin(3x)). As the left hand side is a function of the variable y and the right hand side is a function of the variable x i.e they are implicitly related, we need to use implicit differentiation. This means we differentiate each side of the equation seperately with respect to the functional variable, in this case x. Now differentiate the LHS, which is simple enough
d/dx(y) = dy/dx = f'(x)
To differentiate the RHS, we note that the differential of a logarithm ln(g(x)) is given by g'(x) / g(x), a relationship thatÂ can be found easily by integrating. So in this example we have that g(x) = sin(3x), therefore we need to use the chain rule to differentiate it. The differential of sin(3x) is thus cos(3x) multiplied by the differential of 3x which is 3, therefore g'(x) = 3cos(3x)Â Â and so the differential of the RHS is 3cos(3x)/sin(3x)Â = 3cot(3x)
We thus have dy/dx = 3cot(3x)
see more