Rhiannon T. GCSE Maths tutor, A Level Maths tutor

Rhiannon T.

Unavailable

Mathematics (Bachelors) - Cardiff University

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1 completed lesson

About me

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!

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!

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Personally interviewed by MyTutor

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Ratings & Reviews

0from 1 customer review
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Najma (Parent from Winscombe)

September 27 2016

coudlnt hear her had to cancel

Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A
Further MathematicsA-level (A2)A
Art and DesignA-level (A2)B

General Availability

Pre 12pm12-5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
MathsGCSE£18 /hr

Questions Rhiannon has answered

Find the equation of the tangent to the curve y = 2 ln(2e - x) at the point on the curve where x = e.

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 - y= 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.

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 - y= 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.

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2 years ago

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