Monday, November 30, 2009

Ladder How Fast Is The Top Of The Ladder Approaching The Ground When The Base Is 9 From The Fence?

How fast is the top of the ladder approaching the ground when the base is 9 from the fence? - ladder

A 25-foot ladder rests on a nearly 9 meters. The base of the ladder from the fence to 10 meters per minute. How fast is the top of the ladder, closer to the ground when the base is 9 of the fence? [Note: The scale is over the fence.]

The upper level is closer to the earth of_____ min m / min.

5 comments:

mohanrao d said...

Draw the triangle ABC. \\ \\ \\ \\ \\ \\ \\ \\ U0026lt, B = 90 degrees

AB is around and let y = AB

AC is straight and AC = 25

BC is the distance between the foot of the stairs and railings, is BC = x.

x ^ 2 + y ^ 2 = 25 ^ 2 ------------------( 1)

in a particular case when x = 9 (the base is 9 feet from the fence)

81 + y ^ 2 = 625

y ^ 2 = 544

y = 4 √ 34

Differentiating equation (1) with respect to t

2x dx / dt + 2y dt / dt = 0

x dx / dt + dy / dt = 0

where x = 9, y = 4 √ min 34, dx / dt = 10 m /

9 * (10) √ 34 + 4 dy / dt = 0

4 √ 34 dy / dt = -90

dy / dt = -90 / (4 √ 34)

= -22.5 / √ 34

= -3.86 (Negative because she slipped on the stairs)

The scale slides up to a height of 3.86 m / min

mohanrao d said...

Draw the triangle ABC. \\ \\ \\ \\ \\ \\ \\ \\ U0026lt, B = 90 degrees

AB is around and let y = AB

AC is straight and AC = 25

BC is the distance between the foot of the stairs and railings, is BC = x.

x ^ 2 + y ^ 2 = 25 ^ 2 ------------------( 1)

in a particular case when x = 9 (the base is 9 feet from the fence)

81 + y ^ 2 = 625

y ^ 2 = 544

y = 4 √ 34

Differentiating equation (1) with respect to t

2x dx / dt + 2y dt / dt = 0

x dx / dt + dy / dt = 0

where x = 9, y = 4 √ min 34, dx / dt = 10 m /

9 * (10) √ 34 + 4 dy / dt = 0

4 √ 34 dy / dt = -90

dy / dt = -90 / (4 √ 34)

= -22.5 / √ 34

= -3.86 (Negative because she slipped on the stairs)

The scale slides up to a height of 3.86 m / min

mohanrao d said...

Draw the triangle ABC. \\ \\ \\ \\ \\ \\ \\ \\ U0026lt, B = 90 degrees

AB is around and let y = AB

AC is straight and AC = 25

BC is the distance between the foot of the stairs and railings, is BC = x.

x ^ 2 + y ^ 2 = 25 ^ 2 ------------------( 1)

in a particular case when x = 9 (the base is 9 feet from the fence)

81 + y ^ 2 = 625

y ^ 2 = 544

y = 4 √ 34

Differentiating equation (1) with respect to t

2x dx / dt + 2y dt / dt = 0

x dx / dt + dy / dt = 0

where x = 9, y = 4 √ min 34, dx / dt = 10 m /

9 * (10) √ 34 + 4 dy / dt = 0

4 √ 34 dy / dt = -90

dy / dt = -90 / (4 √ 34)

= -22.5 / √ 34

= -3.86 (Negative because she slipped on the stairs)

The scale slides up to a height of 3.86 m / min

Jose said...

Using similar triangles, we get:

AB = hypotenuse of the triangle of 90 degrees from vertical and horizontal page with X, where 9 is the height of the fence.

small triangle is the triangle of 90 degrees in the major.

25 = hypotenuse of the larger triangle with the vertical part J.

(y / 9) = (25/AB), AB = (x ^ 2 + 9 ^ 2) ^ (1 / 2)


Y = 225 * (x ^ 2 +81) ^ (-1 / 2)

dy / dt = 225 *- 1 / 2 [(x ^ 2 +81) ^ (-3 / 2)] 2xdx/dt ...

Plug-in (dx / dt) = 10 x = 9 in the above equation.

The answer is: [-125 * (2) ^ (1 / 2)] / 18 ft / min

Jose said...

Using similar triangles, we get:

AB = hypotenuse of the triangle of 90 degrees from vertical and horizontal page with X, where 9 is the height of the fence.

small triangle is the triangle of 90 degrees in the major.

25 = hypotenuse of the larger triangle with the vertical part J.

(y / 9) = (25/AB), AB = (x ^ 2 + 9 ^ 2) ^ (1 / 2)


Y = 225 * (x ^ 2 +81) ^ (-1 / 2)

dy / dt = 225 *- 1 / 2 [(x ^ 2 +81) ^ (-3 / 2)] 2xdx/dt ...

Plug-in (dx / dt) = 10 x = 9 in the above equation.

The answer is: [-125 * (2) ^ (1 / 2)] / 18 ft / min

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