Currently unavailable: until 02/02/2017
Degree: Mathematics (Bachelors) - Cardiff University
I am currently in my second year of studying Maths at Cardiff Universty. I love my subject and as a tutor I hope to inspire and instill that love within my students. I'm laid back, friendly, and love to teach. I have been doing Taekwondo for more than 12 years and teaching younger students since the age of 15. This has taught me to be patient and kind, and I take great pride when helping someone overcome any issues. I can't wait to get started with you!
|Maths||A Level||£20 /hr|
|Art and Design||A-Level||B|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Najma (Parent) September 27 2016
The first step is to find the y coordinate at the point on the curve where x = e. To do this we subsitute x = e into the equation of the curve, so y = 2ln(2e - x) becomes y = 2ln(2e - e). Simplifying this, we get y = 2ln(e). Using our knowledge of the natural logarithm, ln(e) = 1, so therefore y = 2.
Next, we find the gradient of the curve at the point x = e. We do this by differentiating the equation, and then subsituting the value for x in. Note that we have to use the chain rule.
dy/dx = -2/(2e - x) = -2/(2e - e) = -2/e.
Using all the information we've found, we can now produce the equation of the tangent line using the equation of a line formula, y - y0 = m(x - x0), where m is the gradient.
y - 2 = -2/e (x - e). Rearranging this, we get y = 4 - 2x/e as the equation of the tangent line.see more