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A circle A has equation x^2+y^2-6x-14y+54=0. Find a) the coordinates of the centre of A, b) the radius of the circle A.

The standard equation of a circle is in the form (x-a)^2+(y-b)^2=r^2, where the coordinates of the centre of the circle is (a,b) and the radius of the circle is r. Therefore, you must put the given equati...

Answered by Evelyn Q. • Maths tutor

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How to do the product rule for differentiation

To do the product rule you must have two functions multiplied together. Here you must make one function be u and the other be v. The formula from this is uv’+vu’ where v’ and u’ are the differentials of t...

Answered by Daisy R. • Maths tutor

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Differentiate xcos(x) with respect to x.

How do we know which method of differentiation to use in this example?
Well in ‘xos(x)’, we have 2 different functions: ‘x’ and ‘cos(x)’. Therefore, we must differentiate using the p...

Answered by Oliver S. • Maths tutor

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A particle P is projected vertically upwards from a point 20m above the ground with velocity 18m/s, no external forces act on it other than gravity. What will its speed be right before it hits the ground? Give your answer to one decimal place.

To start off we should list what we have and what we want to find. The initial velocity u= 18 m/s (taking upwards to be the positive direction). Acceleration a= - 9.8 m/s² (negative since gravi...

Answered by Jenny N. • Maths tutor

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The curve C has equation y=(2x-3)^5, the point P lies on C and has coordinates (w, – 32), find (a) the value of w and (b) the equation of the tangent to C at the point P in the form y=mx+c , where m and c are constants.

(a) The curve is defined by y=(2x-3)^5. To find x=w when y=-32, we must substitute these values into the equation C and re-arrange to find w. -32=(2w-3)^5. First we must remove the power of 5 by doing pow...

Answered by Jordan P. • Maths tutor

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