022 Exam 1 Sample A, Problem 6 - Revision history

MathAdmin at 05:43, 13 April 2015

2015-04-13T05:43:24Z

MathAdmin at 05:40, 13 April 2015

2015-04-13T05:40:11Z

MathAdmin at 05:39, 13 April 2015

2015-04-13T05:39:30Z

MathAdmin: Created page with "350px A 15-foot ladder is leaning against a house. The base of the ladder is pulled away from the house at a rate of 2 feet p..."

2015-04-13T05:39:08Z

Created page with "right|350px <span class="exam">A 15-foot ladder is leaning against a house. The base of the ladder is pulled away from the house at a rate of 2 feet p..."

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[[File:022_S1_A_6.png|right|350px]]

<span class="exam">A 15-foot ladder is leaning against a house. The base of
the ladder is pulled away from the house at a rate of 2 feet per second.
How fast is the top of the ladder moving down the wall when the base
of the ladder is 9 feet from the house.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
|-
|Like most geometric word problems, you should start with a picture. This will help you declare variables and write meaningful equation(s). In this case, we will have to use implicit differentiation to arrive at our related rate.
|}
 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|'''Write the Basic Equation:''' From the picture, we can see that the ladder forms a right triangle with the wall and the ground, so we can treat our variables as
|-
|
::<math>x^2+y^2\,\,=\,\,15^2\,\,=\,\,225,</math>
|-
|where <math style="vertical-align: 0%">x</math> is the height of the ladder on the wall, and <math style="vertical-align: -20%">y</math> is the distance between the wall and the base of the ladder.
|}
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|'''Use Implicit Differentiation:''' We take the derivative of the equation from Step 1 to find
|-
|
:: <math>2x\frac{dx}{dt} + 2y\frac{dy}{dt}\,\,=\,\,0,</math>
|-
|or
|-
|
::<math>\frac{dy}{dt} \,\,=\,\, -\frac{x}{y}\cdot\frac{dx}{dt},</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|'''Evaluate and Solve:''' At the particular moment we care about,
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|
::<math>x=9,\quad y=12,\quad dx/dt=2.</math>
|-
|From this, we can simply plug in to find
|-
|
::<math>\frac{dy}{dt} \,\, = \,\, - \frac {x}{y} \cdot \frac{dx}{dt} \,\, = \,\, -\frac{9}{12} \cdot 2 \,\, = \,\, - \frac{3}{2}</math>
|-
|With units, we have that the ladder is sliding down the wall at <math style="vertical-align: -25%">-3/2</math>  feet per second.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|With units, we have that the ladder is sliding down the wall at <math style="vertical-align: -25%">-3/2</math>  feet per second.
|}

[[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 4: / Line 4: @@
 the ladder is pulled away from the house at a rate of 2 feet per second.
 How fast is the top of the ladder moving down the wall when the base
-of the ladder is 9 feet from the house.
+of the ladder is 9 feet from the house?
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 21: / Line 21: @@
 ::<math>x^2+y^2\,\,=\,\,15^2\,\,=\,\,225,</math>
 |-
-|where <math style="vertical-align: 0%">x</math> is the height of the ladder on the wall, and <math style="vertical-align: -20%">y</math> is the distance between the wall and the base of the ladder.
+|where <math style="vertical-align: 0%">x</math> is the distance between the wall and the base of the ladder, and <math style="vertical-align: -20%">y</math> is the height of the ladder on the wall.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 51: / Line 51: @@
 |-
 |With units, we have that the ladder is sliding down the wall at <math style="vertical-align: -25%">-3/2</math>&thinsp; feet per second.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-