一條問題 - Applications of Differentiation : Rate of Change

oscar

In the figure, ABCD is a square of side a cm. P is a variable point on AB such that angle ADP = α and angle BCP = θ. If θ increases at a rate of 1 rad./sec when θ = π/6, find the rate of change of α.

(Copied from New Progress in Additional Mathematics 3 Exercise 15.5 Question 17)

抹茶可樂

(First we find out the relationship between αandθ)
AP+PB=AB
ADtanα+BCtanθ=AB
Since AD=BC=AB
tanα+tanθ=1
(It implies that the sizes of angles are indepedent to the length of square)

When θ=π/6, PB=atanπ/6=√3a/3
tanα=AP/AD=1-1/√3

(secα)^2．dα/dt+(secθ)^2．dθ/dt=0
[1+(tanα)^2]dα/dt+[1+(tanθ)^2]dθ/dt=0
[1+(1-1/√3)^2]dα/dt+[1+1/3](1)=0
(7/3-2/√3)dα/dt=-4/3
dα/dt=-4/(7-2√3) rad/sec

(α is decresing)

[ Last edited by 落雷 on 2005-7-27 at 08:40 PM ]

oscar

Originally posted by 落雷 at 2005-7-27 20:36:
(First we find out the relationship between αandθ)
AP+PB=AB
ADtanα+BCtanθ=AB
Since AD=BC=AB
tanα+tanθ=1
(It implies that the sizes of angles are indepedent to the length of square)

Wh ...

係喎...
我唔記得將sec^2 α轉做1+tan^2 α添...
唔該晒！[E-Hap]

細薯

我唔識ge．．>_<
CE ga﹖

oscar

Originally posted by smallpotato226 at 2005-7-29 13:04:
我唔識ge．．>_<
CE ga﹖

當然啦...
不過係中五先教....
我都係學定先je...

JOHNLAM17

Originally posted by oscar at 2005-7-29 10:03 PM:


當然啦...
不過係中五先教....
我都係學定先je...

你好勁ar....
我一望到a.maths第一課(mathematical induction<--唔記得左個名)就唔想再望=.=
不過果d 唔知咩(q字頭的)equations 就易d

oscar

Originally posted by JOHNLAM17 at 2005-7-29 22:53:

你好勁ar....
我一望到a.maths第一課(mathematical induction<--唔記得左個名)就唔想再望=.=
不過果d 唔知咩(q字頭的)equations 就易d

第一課係MI？唔係Quadratic Equations and Absolute Values咩？
其實MI都係prove je...
學學下就會覺好easy
Quadratic Equations係A Maths最容易...唔識就...

細薯

MI 最易 bor....
唔使溫ge....
(個人覺得)

oscar

Originally posted by smallpotato226 at 2005-7-31 12:53:
MI 最易 bor....
唔使溫ge....
(個人覺得)

我覺得最易係Quadratic、MI同Binomial
最難係Coordinate同Circle...(因為成日都唔記得d formulae...)

丫奴

Originally posted by smallpotato226 at 2005-7-29 01:04 PM:
我唔識ge．．>_<
CE ga﹖

我地學校中四已經教ｒａｔｅ　ｏｆ　ｃｈａｎｇｅ．．．．．