MATH P1 Help!!!!

[Nikesh]

The Equation of curve is y=root(5x+4).
i) Calculate the gradient of the curve at the point where x=1.
ii) A point with coordinates (x,y) moves along the curve in such a way that the rate of increase of x has the constant value 0.03 units per second. Find the rate of increase of y at the instant when x=1.
iii)Find the area enclosed by the curve, the x-axis, the y-axis and the line x=1.

 

[OakMoon!]

It's simple calculus.
i) Find the dericative of y with respect to x. d/dx[square root(5x+4)]= 5/(2*square root(5x+4))... Put in 1 and you get gradient=5/6
ii) Put in the chain rule. dy/dt=dy/dx*dx/dt
so dy/dt=5/6*0.03=1/40
iii)Find the integral with the limits 0 to 1..
Which is [2(5x+4)^(3/2)]/15
Put in the limits answer is 3.6-1.1=2.5