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We first need to find dy/dx and we use the fact that dy/dx = dy/dt * dt/dx. So we have dy/dt = -6sin(3t) and dx/dt = 12cos(3t). Substituing these in we have dy/dx = -6*sin(3t)1/(12cos(3t...

To find the turning points we need to find when the differential of the equations with respect to x is equal to 0. (dy/dx = 3x² - 12 = 0) From this we find that the turning points happ...

To solve such equations we take advantage of log lawes to simplify the problem .

E.g

ln[sqrt(1-x²)] = ln[(1-x²)^1/2] = 1/2ln[1-x²]

After sim...

To solve this you need to integrate by substitution. You can spot this because the differential of the bottom of the fraction is a multiple of the top part, showing this quickly; if u = x²  + 3...

The equation we use to integrate by parts is

y = uv - ∫ v(du/dx) dx + c

so we separate dy/dx into u=(3x-4) and dv/dx=(2x²+5)

however we still need to find du/dx an...