Rates of change (1 Viewer)

[taggs-sasuke]

I need help with this question:

A ladder AB, 5m long is leaning against a vertical wall with its foot on horizontal ground OB. The foot of the ladder begins to slide along the ground away from the wall at a constant speed of 1 m/s.

Find the speed at which the top of the ladder AB is moving down the wall at the time when the ladder is 3m from the wall.

Answer:

-3/4 m/s

Thank you!!!

Edit: Sorry, I think this is a 2U question.

 

[GUSSSSSSSSSSSSS]

let the vertical height of the ladder = x
let the horizontal length of the ladder out from the wall = y

using pythagoras: x^2 + y^2 = 5^2 = 25
y^2 = 25 - x^2
y = sqrt(25 - x^2) (since distances are measured as positive quantities it is the positive sqrt)

rate at which it slips from the top is dx/dt
dx/dt = dx/dy . dy/dt (chain rule)

question gives that dy/dt = 1
therefore: dx/dt = dx/dy

Now dy/dx = (-x/sqrt(25 - x^2))
therefore: dx/dy = -sqrt(25 - x^2)/x
therefore: dx/dt = -sqrt(25 - x^2)/x

it is when the ladder is 3m from the wall....ie: y = 3
x^2 + 3^2 = 25
x^2 = 16
x = 4 (once again it is ONLY positive)

therefore: dx/dt = -sqrt(25 - 4^2)/4
= -sqrt(9)/4
= -3/4

 

[taggs-sasuke]

Thank you Gus!


Sadly, I need help again!


Can someone please show me how to do the following questions? (Thank you!)


1.

In triangle ABC, AB = 10 cm, AC = 12 cm and angle A is increasing at the rate of o.1 radian/second. At what rate is:

(a) the area of triangleABC increasing,

(b) the length of BC increasing, when angleA is pie/3 radians?

Answer:

(a) 3 cm^2/s
(b) 3 square root(3/31) cm/s


2.

A loading chute is in the shape of a right square pyramid of base length 10m and depth 8m. Liquid is poured in the top at a rate of 4 m^3/min. At what rate is the level rising when the depth is 4m?

Answer:

4/25 m/min


Thank you very much!!!

 

[gurmies]

http://community.boredofstudies.org...282/applications-calculus-physical-world.html

Answers to your questions taggs =]

 

[taggs-sasuke]

http://community.boredofstudies.org...282/applications-calculus-physical-world.html

Answers to your questions taggs =]

Oh! Thanks!

 

[gurmies]

Oh! Thanks!

No worries =]

 

[taggs-sasuke]

Another question:

6. Show that the rate of change of the area of a circle with respect to the radius is proportional to the radius.

Source: Fitzpatrick 2U

Thank you!

 

[study-freak]

 

[gurmies]

To explain, dA/dr = 2pir, = Kr, where K is a constant. Essentially that's what proportionality is =]

 

[taggs-sasuke]

I get it. Thank you both.


Another question:

Fitzpatrick 3U

25) A point P moves on the curve y=x^2 and the x-coordinate increases at a constant rate of 5 units per second. At what rate is:

(a) the y coordinate of P increasing when x = 1?
(b) the gradient of the curve increasing when x = 1?



I still don't understand part (b).

Can someone please explain it to me? (Thank you!)

 

[GUSSSSSSSSSSSSS]

ahhhhhh k

the gradient is dy/dx

therfore the rate of change of the gradient is: (dy/dx)/dt

using chain rule = (dy/dx)/dx . dx/dt

(dy/dx)/dx is the differential of dy/dx with respect to x ..... ie: the 2nd derivative

(dy/dx)/dx = 6x

therefore: (dy/dx)/dt = 6x . 5

measured when x = 1

therefore (dy/dx)/dt = 6(5) = 30




is that better???

 

[taggs-sasuke]

Another question:

Question in red.

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Answer: 4/25 m/min

dr/dt = dv/dt x dr/dv
dv/dt = 4
so we need to find dr/dv

B/h = 10/8
B = 5h/4

V = 1/3.B^2H - Is this the formula for the volume of a (right) square pyramid?
but B = 5h/4
V = 1/3.(5h/4)^2.h
= 1/3.(25h^2/16).h
= (25h^3)/48
dv/dr = (75h^2)/48

dr/dt = dv/dt x dr/dv
= 4 x 48/(75h^2)
when h =4
= 4 x 48/1200
= 48/300
= 4/25 m/min

 

[taggs-sasuke]

ahhhhhh k

the gradient is dy/dx

therfore the rate of change of the gradient is: (dy/dx)/dt

using chain rule = (dy/dx)/dx . dx/dt

(dy/dx)/dx is the differential of dy/dx with respect to x ..... ie: the 2nd derivative

(dy/dx)/dx = 6x

therefore: (dy/dx)/dt = 6x . 5

measured when x = 1

therefore (dy/dx)/dt = 6(5) = 30




is that better???

Yes, I get it now. YAY!

Thank you!

 

[GUSSSSSSSSSSSSS]

yea the formula for a square pyramid is: 1/3 of the volume of the rectangle which is made of that base and height

ie: (1/3)Base.Height

 

[taggs-sasuke]

yea the formula for a square pyramid is: 1/3 of the volume of the rectangle which is made of that base and height

ie: (1/3)Base.Height

Ohhh. Yes, I read it wrong. I thought it was b to the power of 2h instead of b squared times h.

Yay, it's all good now. Thanks.