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To find the gradient of the curve at t=2 we need to find an expression for dy/dx and then substitute in for t=2. We can make use of the chain rule to find this expression because dy/dx = (dy/dt)/(dx/dt) a...

First identify that integration by parts is required. Then seperate the integration so u = ln(x)     dv/dx = x then, du/dx = 1/x  v = (1/2)x^2 . And using the integration by parts formula with these subst...

y is a function of x₁ and x₂. We are asked to derive y with respect to x₁, meaning that x₂ remains constant. 

Note that y' is the deriva...

First we need to change the limits, and by plugging in 1 and 0 to our substitution we find that the limits for u are arctan(1) and arctan(0) (or pi/4 and 0). Then we need to substitute for dx, and by diff...

M=37.5e^kt 52=37.5e^k5 52/37.5=e^5k ln(52/37.5)=5k (1/5)(ln(52/37.5))=k k≈0.06538 when t=9, M=37.5e^9*0.06538 M=67.54