Answer

Question

A ladder 10 m long leans against a vertical wall. The bottom is pulled away at a rate of 1 m/s.

How fast is the top sliding down the wall when the bottom is 6 m from the wall?

Answer

the length of the ladder l = 10 m, 
the distance from the bottom of the ladder to the wall x = 6 m
the height of the top of the ladder on the wall = y 
using the pythagorean theorem :
y^2 = l^2 - x^2 = 10^2 - 6^2 = 100 - 36 = 64
y = 8 m
the bottom is pulled away at a rate of 1 m/s
the distance the bottom pulled away through the top end touch the ground  is 10 m - 6 m = 4 m
 total time for the bottom pulled away through the top end touch the ground = 4 m / 1 m/s = 4 s
simultaneously, total time for the top of ladder to touch the ground is 4 s
so,the rate of the top sliding down the wall = 8 m / 4 s = 2 m/s

Answer

we are given a 10 m ladder leaning against a vertical wall. let x be the distance of the bottom of the ladder from the wall and y the height of the ladder on the wall. the relationship is: x^2 + y^2 = 100 differentiating with respect to time t: 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 simplify: \frac{dy}{dt} = -\frac{x}{y} \cdot \frac{dx}{dt} when the bottom is 6 m from the wall, x = 6 \quad ightarrow \quad y = \sqrt{100 - 36} = 8 substitute values: \frac{dy}{dt} = -\frac{6}{8} \cdot 1 = -0.75 \, ext{m/s} thus, the top of the ladder slides down the wall at a rate of 0.75 m/s when the bottom is 6 m away. the negative sign confirms that the height is decreasing. final answer: the top slides down at 0.75 m/s.

Answer

let

x

be the bottom distance from the wall and

y

the height on the wall. ladder is constant:

x^2 + y^2 = 10^2

. differentiate:

y˙=0⇒y˙=−xyx˙2x\,\dot x + 2y\,\dot y = 0 ightarrow \dot y = -\frac{x}{y}\dot x

. given

m/s\dot x = 1\ ext{m/s}

and

x = 6

. then

y=100−62=64=8y=\sqrt{100-6^2}=\sqrt{64}=8

=64 =8. so

m/s\dot y = -\frac{6}{8}\cdot 1 = - frac{3}{4}\ ext{m/s}

. answer: the top slides down at

m/s0.75\, ext{m/s}

(negative sign just means downward).

Answer

Answer

step-by-step solution with explanation
step 1: define the variables
let:
(x) = distance of the bottom of the ladder from the wall (in meters)(y) = height of the top of the ladder above the ground (in meters)
the ladder has a fixed length of 10 m, so it forms a right triangle.
step 2: write the relationship between (x) and (y)
since the ladder, wall, and ground form a right triangle, use the pythagorean theorem:
[
x^2 + y^2 = 10^2
]
simplify:
[
x^2 + y^2 = 100
]
this equation relates the two changing distances.
step 3: differentiate with respect to time
both (x) and (y) change over time, so differentiate every term with respect to time (t).
differentiate:
[
\frac{d}{dt}(x^2+y^2)=\frac{d}{dt}(100)
]
using the chain rule:
the derivative of (x^2) is (2x\dfrac{dx}{dt}).the derivative of (y^2) is (2y\dfrac{dy}{dt}).the derivative of the constant (100) is (0).
therefore,
[
2x\frac{dx}{dt}+2y\frac{dy}{dt}=0
]
divide every term by 2:
[
x\frac{dx}{dt}+y\frac{dy}{dt}=0
]
this is the equation we will use to find the required rate.
step 4: write down the given information
the problem states:
the bottom is pulled away from the wall at 1 m/s.
therefore,
[
\frac{dx}{dt}=1\ 	ext{m/s}
]
when the bottom is 6 m from the wall,
[
x=6\ 	ext{m}
]
we still need to find (y).
step 5: find the height of the ladder
use the pythagorean equation:
[
x^2+y^2=100
]
substitute (x=6):
[
6^2+y^2=100
]
[
36+y^2=100
]
subtract 36 from both sides:
[
y^2=64
]
take the positive square root (since height cannot be negative):
[
y=8\ 	ext{m}
]
step 6: substitute into the differentiated equation
recall:
[
x\frac{dx}{dt}+y\frac{dy}{dt}=0
]
substitute:
(x=6)(\dfrac{dx}{dt}=1)(y=8)
[
6(1)+8\frac{dy}{dt}=0
]
simplify:
[
6+8\frac{dy}{dt}=0
]
subtract 6 from both sides:
[
8\frac{dy}{dt}=-6
]
divide both sides by 8:
[
\frac{dy}{dt}=-\frac{6}{8}
]
reduce the fraction:
[
\frac{dy}{dt}=-\frac{3}{4}\ 	ext{m/s}
]
or
[
\frac{dy}{dt}=-0.75\ 	ext{m/s}
]
final answer
[
\boxed{\frac{dy}{dt}=-\frac{3}{4}\ 	ext{m/s}}
]
interpretation
the negative sign indicates that the height (y) is decreasing. therefore, when the bottom of the ladder is 6 m from the wall, the top is sliding down the wall at a rate of (\boxed{0.75\ 	ext{m/s}}).

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