An 13'ladder is sliding down a wall. If the distancebetween the top of the ladder and the floor isdecreasing at a rate of 3 feet/second when theladder is 12' above the floor, determine the rate atwhich the base of the ladder is moving away fromthe wall.dxNote, you are trying to finddtx'(t).13'y(t)x(t)(a) Determine x =x(t) wheny= y(t)= 12.x =feet.(b) Write an equation in t relating the values on thetriangle.Equation in tImplicitly differentiate the equation with respect todxat the prescribed moment.dtt and determine(c) The base of the ladder is moving away from thedxwall atdtfeet/second.

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Description: An 13'ladder is sliding down a wall. If the distancebetween the top of the ladder and the floor isdecreasing at a rate of 3 feet/second when theladder is 12' above the floor, determine the rate atwhich the base of the ladder is moving away fromthe w

From the given problem: dydt=-3 feet/sec & y=12' From the given figure:…

Answered: An 13 'ladder is sliding down a wall.…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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An 13'ladder is sliding down a wall. If the distance
between the top of the ladder and the floor is
decreasing at a rate of 3 feet/second when the
ladder is 12' above the floor, determine the rate at
which the base of the ladder is moving away from
the wall.
dx
Note, you are trying to find
dt
x'(t).
13'
y(t)
x(t)
(a) Determine x =
x(t) when
y
= y(t)
= 12.
x =
feet.
(b) Write an equation in t relating the values on the
triangle.
Equation in t
Implicitly differentiate the equation with respect to
dx
at the prescribed moment.
dt
t and determine
(c) The base of the ladder is moving away from the
dx
wall at
dt
feet/second.

Expert Solution

Step 1

From the given problem:

$\frac{d y}{d t} = - 3 f e e t / s e c & y = 12'$

From the given figure:

$\begin{array}{rcl} P e r p e n d i c u l a r & = & y (t) \\ H y p o t e n u s e & = & 13' \\ B a s e & = & x (t) \end{array}$

Given:

$y = y (t) = 12$

Applying the Pythagoras theorem formula:

${(H y p o t e n u s e)}^{2} = {(P e r p e n d i c u l a r)}^{2} + {(B a s e)}^{2}$

So,

$\begin{array}{rcl} {(x (t))}^{2} + {(y (t))}^{2} & = & {(13')}^{2} . . . (1) \\ {(x (t))}^{2} + y^{2} & = & 13^{2} \\ {(x (t))}^{2} + {(12)}^{2} & = & 169 \\ {(x (t))}^{2} + 144 & = & 169 \\ {(x (t))}^{2} & = & 169 - 144 \\ x (t) & = & 5 f e e t \end{array}$

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