Top answers

Further Mathematics
All levels

y = artanh(x/sqrt(1+x^2)) , find dy/dx

Some of these examiners quite like asking students to find the derivative of an inverse trig or hyperbolic function to try and catch someone off guard. The best way to approach these is to first multiply ...

Answered by Eugene L. Further Mathematics tutor
5466 Views

Show that the sum from 1 to n of 1/(2n+1)(2n-1) is equal to n/(2n+1) by Induction

First we check that this is true for n=1: S1 = 1/(1x3)  which is equal to n/(2n+1) for n=1 therefore Sn = n/(2n+1) is true for n = 1. Next assume that it is true for n=k. Sk

Answered by Jamie F. Further Mathematics tutor
12933 Views

Two planes have eqns r.(3i – 4j + 2k) = 5 and r = λ (2i + j + 5k) + μ(i – j – 2k), where λ and μ are scalar parameters. Find the acute angle between the planes, giving your answer to the nearest degree.

Summary of solution: To find the angle between the planes, we must find the normal vector to each plane and then use the scalar product to find the angle between these two normal vectors....

Answered by Daniel C. Further Mathematics tutor
6992 Views

Find the four complex roots of the equation z^4 = 8(3^0.5+i) in the form z = re^(i*theta)

We know that z=re^(itheta) from the definition of the exponential form of a complex number. Hence it follows that: z^4=(re^(itheta))^4=r^4e^(4itheta) We can find z^4 by converting 8(...

Answered by George G. Further Mathematics tutor
5391 Views

Find, without using a calculator, integral of 1/sqrt(15+2x-x^2) dx, between 3 and 5, giving your answer as a multiple of pi

To get the denominator into something usable, you have to complete the square so you have it in one of the forms you can use a trig or hyperbolic substitution for. The minus sign in front of the x2 means ...

Answered by Luke D. Further Mathematics tutor
3944 Views

We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences