# applications of differentiation #2

• Jun 19th 2010, 11:56 AM
Shiraz
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!
• Jun 26th 2010, 05:28 AM
CaptainBlack
That is incomprehensible, try rewording and/or including any missing information.

CB