🔒 Closed Avengers end game 1 ticket contest real quick #2

[cleanfuel]

Rules:

1. Paunahan lang
2. Show your solution
3. Hangang 7pm lang 04/25/2019 GMT +8

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Question:

A 15 foot ladder is resting against the wall. The bottom is intially 10 feet away from the wall and is being pushed towards the wall at the rate of 1/4 ft/sec. How fast is the top of the ladder moving up the wall 12 seconds after we start pushing?

 

[Sazuree8]

Pythagorean theorem.
x2+y2=(15)2=225
differentiate both sides with respect to tt, remembering that xx and yy are really functions of t and so we’ll need to do implicit differentiation.
2xx′+2yy′=0
Determining xx and yy is actually fairly simple. We know that initially x=10x=10 and the end is being pushed in towards the wall at a rate of 1414ft/sec and that we are interested in what has happened after 12 seconds. We know that,
distance=rate×time=(14)(12)=3
So, the end of the ladder has been pushed in 3 feet and so after 12 seconds we must have x=7x=7. Note that we could have computed this in one step as follows,
x=10−14(12)=7
To find yy (after 12 seconds) all that we need to do is reuse the Pythagorean Theorem with the values of xx that we just found above.
y=√225−x2=√225−49=√176
Now all that we need to do is plug into You do not have permission to view the full content of this post. Log in or register now.(1) and solve for y′
2(7)(−14)+2(√176)y′=0⇒y′=7/4√176=74√176=0.1319ft/sec

 

[cleanfuel]

x2+y2=(15)2=225
2xx′+2yy′=0
distance=rate×time=(14)(12)=3
x=10−14(12)=7
y=√225−x2=√225−49=√176
2(7)(−14)+2(√176)y′=0⇒y′=7/4√176=74√176=0.1319ft/sec

pa step by step sir

 

[- Glyph -]

The length of the ladder is I.

The top of the ladder is at a distance of y from the ground.

The bottom of the ladder be at a distance of x from the wall

We have to know how fast is the top of the ladder moving up the wall 12 seconds after we start pushing.

From Pythagoras theorem

l2=x2+y2l2=x2+y2

Differentiate both sides with respec to tt which impllesddt
(l2)=ddt(x2+y2)⇒0=2xdxdt+2ydydtdydt=(−xy)dxdy

(dxdt)=0.25ft/secd=(dxdt)×t=0.25×12=3(dxdt)=0.25ft/secd=(dxdt)×t=0.25×12=3

Therefore, the values of x from the wall is

X=initial distance−3=10−3=7X=initial distance−3=10−3=7

Calculate the values of y at t=12 secy at t=12 sec given x=7 and l=15x=7 and l=15 implles

152=72+y2⇒y=√225−49=√176=4√11152=72+y2⇒y=225−49=176=411

Therefore, the rate at which the ladder is moving up when t=12 sect=12 sec

dydt|t=12=|(−xy)t=12(dxdt)|=74√11×(0.25)=7√11176=0.1319ft/sec

 

[Assmodeus]

N

The length of the ladder is I.

The top of the ladder is at a distance of y from the ground.

The bottom of the ladder be at a distance of x from the wall

We have to know how fast is the top of the ladder moving up the wall 12 seconds after we start pushing.

From Pythagoras theorem

l2=x2+y2l2=x2+y2

Differentiate both sides with respec to tt which impllesddt
(l2)=ddt(x2+y2)⇒0=2xdxdt+2ydydtdydt=(−xy)dxdy

(dxdt)=0.25ft/secd=(dxdt)×t=0.25×12=3(dxdt)=0.25ft/secd=(dxdt)×t=0.25×12=3

Therefore, the values of x from the wall is

X=initial distance−3=10−3=7X=initial distance−3=10−3=7

Calculate the values of y at t=12 secy at t=12 sec given x=7 and l=15x=7 and l=15 implles

152=72+y2⇒y=√225−49=√176=4√11152=72+y2⇒y=225−49=176=411

Therefore, the rate at which the ladder is moving up when t=12 sect=12 sec

dydt|t=12=|(−xy)t=12(dxdt)|=74√11×(0.25)=7√11176=0.1319ft/sec

Nakita ko din yan sa google zer

 

[Sazuree8]

pa step by step sir

Lalagyan ko ng explanation sir?

 

[lllumi]

Idol talaga

 

[- Glyph -]

N

Nakita ko din yan sa google zer

Walang rules na d pwede i google

 

[Assmodeus]

Walang rules na d pwede i google

Yep tama yan

 

[lllumi]

Walang rules na d pwede i google

Lol nakita ko rin yan

 

[Anonomous]

Given
(dx/dt)=0.25ft/sec
d=(dx/dt)×t=0.25×12=3
Therefore, the values of x from the wall is
X=initial distance−3=10−3=7
Calculate the values of y at t=12 secy at t= 12 sec given x= 7 and l= implles
15^2=7^2+y^2⇒y=√225−49=√176=4√11
Therefore, the rate at which the ladder is moving up when t=12 sect=12 sec
dy/dt|t = 12 = |(−x/y)t = 12(dx/dt)|
=74√11×(0.25)
=7√11176
=0.1319ft/sec

Gensan haha

 

[lllumi]

Sayang math eh kung about mcu to baka may chance pa

 

[Assmodeus]

G

Given
(dx/dt)=0.25ft/sec
d=(dx/dt)×t=0.25×12=3
Therefore, the values of x from the wall is
X=initial distance−3=10−3=7
Calculate the values of y at t=12 secy at t= 12 sec given x= 7 and l= implles
15^2=7^2+y^2⇒y=√225−49=√176=4√11
Therefore, the rate at which the ladder is moving up when t=12 sect=12 sec
dy/dt|t = 12 = |(−x/y)t = 12(dx/dt)|
=74√11×(0.25)
=7√11176
=0.1319ft/sec

Gensan haha

Google din

 

[Anonomous]

G

Google din

atleast inayos ko haha

 

[Tilamsik]

dapat related sa movie yung question

 

[lllumi]

dapat related sa môviê yung question

I second this para sa legit fan ng mcu mapunta. Sulit yun

 

[onetoughguy]

Given
(dx/dt)=0.25ft/sec
d=(dx/dt)×t=0.25×12=3
Therefore, the values of x from the wall is
X=initial distance−3=10−3=7
Calculate the values of y at t=12 secy at t= 12 sec given x= 7 and l= implles
15^2=7^2+y^2⇒y=√225−49=√176=4√11
Therefore, the rate at which the ladder is moving up when t=12 sect=12 sec
dy/dt|t = 12 = |(−x/y)t = 12(dx/dt)|
=74√11×(0.25)
=7√11176
=0.1319ft/sec

Gensan haha

gensan engineering student? hahaha

 

[Anonomous]

gensan engineering student? hahaha

oo bagong graduate

 

[Sazuree8]

Sino po ang nanalo"?