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applications of differentiation #2

The diagram below is of two parabola (ASD and BTD) joined together smoothly.

a) Find the value of y1.

Consider the points (27.5, 19.5)and (27,y1)

$\displaystyle \frac{27}{27.5} = \frac{y}{19.5}$

$\displaystyle y1 = \frac{1053}{55} = 19\frac{8}{55}$

Is this correct? Sorry don't know how to use latex to make it all look neat.

b) Parabola BTD contains the point D and has a zero gradient. Show that its equation is $\displaystyle y = \frac{-2}{99}(x-60)^2 + 40$

c) Use calculus to find the gradient of the parabola BTD at point 0.

d) Hence find the equation of the parabola ASD containing point A and joined smoothly to parabola BTD at point D. Curves joined smoothly at a point have the same gradient that that point.

Please keep any working out as simple as possible so that I can understand. Any help regarding these questions will be appreciated!