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Maths
A Level

June 2008 C1 Paper Differentiation Question

a) ask student: purpose of differentiation (i), how to perform differentiation (ii) (i) to find the gradient of a curve (ii) multiply the coefficient of x by its indice then minus one from the indici perf...

Answered by Nicola G. Maths tutor
3142 Views

Using the result: ∫(2xsin(x)cos(x))dx = -1⁄2[xcos(2x)-1⁄2sin(2x)] calculate ∫sin²(x) dx using integration by parts

Recall that ∫uv'=uv- ∫u'v Set u=sin²(x), v'=1 Therefore u'=2sin(x)cos(x) and v=x which gives us the following:

∫sin²(x)dx = xsin²(x) - ∫2xsin(x)cos(x)dx

The second integral in the above expr...

Answered by Nick M. Maths tutor
5042 Views

solve 2cos^2(x) - cos(x) = 0 on the interval 0<=x < 180

we start  y factoring and solving for each equation:

cos(x) (2cos(x) - 1) = 0 

this means: 

cos(x) = 0 and cos(x) = 1/2

from the first equation we get:   x = 90

and from...

Answered by Dimitris S. Maths tutor
7837 Views

Express 2cos(x) + 5sin(x) in the form Rsin(x + a) where 0<a<90

Expanding Rsin(x + a): Rsin(x + a) = Rsin(x)cos(a) + Rcos(x)sin(a) Comparing coefficients of sin(x), cos(x) with first expression leads to: Rsin(a) = 2, Rcos(a) = 5 Dividing these equations gives: tan(a) ...

Answered by Dan H. Maths tutor
10087 Views

A girl kicks a ball at a horizontal speed of 15ms^1 off of a ledge 20m above the ground. What is the horizontal displacement of the ball when it hits the ground?

As we are looking for the horizontal displacement first we look at horizontal motion. We know that the horizontal velocity is 15ms^-1 but we dont know the time so we can't work out the horizontal displace...

Answered by Victoria W. Maths tutor
3488 Views

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