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The curve C has the parametric equations x=4t+3 and y+ 4t +8 +5/(2t). Find the value of dy/dx at the point on curve C where t=2.

a) What can we find from what we have been given? dx/dt and dy/dt How can we relate these values to dy/dx? In the context of equations that only contain two variables, their derivatives behave like fractions...
CB
Answered by Chloe B. • Maths tutor
10661 Views

Starting from the fact that acceleration is the differential of velocity (dv/dt = a) derive the SUVAT equations.

Intergrating with respect to time, you get that v = u + at. Knowing that velocity is just the rate of change of your position ds/dt = v, and sustituting the previous expression for v, you get ds/dt = u + at....
BW
Answered by Ben W. • Maths tutor
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Using integration by parts, and given f(x) = 3xcos(x), find integrate(f(x) dx) between (pi/2) and 0.

We begin by quoting the integration by parts formula, as the question speciaficaly asks us to use it. integrate(u(x) v'(x) dx)|^(b) (a) = [u(x) v(x)]^(b) (a) - integrate(u'(x) v(x) dx)|^(b)_(a) To use this f...
AC
Answered by Aaron C. • Maths tutor
4190 Views

integrate by parts ln(x)/x^3

The question states to use integration by parts. So first we recall the integration by parts formula is integrate(u(x)v'(x) dx)= (v(x)u(x)) - integrate(u'(x)v(x) dx)+c (note these integrals are with respect ...
PS
Answered by Prit S. • Maths tutor
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Given that the equation of the curve y=f(x) passes through the point (-1,0), find f(x) when f'(x)= 12x^2 - 8x +1

Firstly, Integrate the f'(x) equation by raising the power by 1 and then dividing by the new power and adding a constant c. This gives you f(x)=(12x^3)/3 -(8x^2)/2 + x + c Then you simplify, f(x)=4x^3 -4x^2 ...
DM
Answered by Daniel M. • Maths tutor
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