Difference between revisions of "009A Sample Final 1, Problem 5"

From Grad Wiki
Jump to navigation Jump to search
Line 31: Line 31:
 
|So, we have <math style="vertical-align: -5px">2(40)6=2(50)s'</math>.
 
|So, we have <math style="vertical-align: -5px">2(40)6=2(50)s'</math>.
 
|-
 
|-
|Solving for <math style="vertical-align: 0px">s'</math>, we get <math style="vertical-align: -14px">s'=\frac{24}{5} </math>m/s.
+
|Solving for <math style="vertical-align: 0px">s'</math>, we get <math style="vertical-align: -14px">s'=\frac{24}{5}</math> m/s.
 
|}
 
|}
  
Line 37: Line 37:
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
| <math>s'=\frac{24}{5} </math>m/s
+
| <math>s'=\frac{24}{5}</math> m/s
 
|}
 
|}
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 11:29, 24 February 2016

A kite 30 (meters) above the ground moves horizontally at a speed of 6 (m/s). At what rate is the length of the string increasing

when 50 (meters) of the string has been let out?

Foundations:  

Solution:

Step 1:  
Insert diagram.
From the diagram, we have by the Pythagorean Theorem.
Taking derivatives, we get
.
Step 2:  
If , then .
So, we have .
Solving for , we get m/s.
Final Answer:  
m/s

Return to Sample Exam