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

Revision as of 13:16, 18 March 2017

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:
The Pythagorean Theorem
For a right triangle with side lengths $a,b,c$ where $c$ is the length of the
hypotenuse, we have $a^{2}+b^{2}=c^{2}.$

Solution:

Step 1:
Insert diagram.
From the diagram, we have $30^{2}+h^{2}=s^{2}$ by the Pythagorean Theorem.
Taking derivatives, we get
$2hh'=2ss'.$

Step 2:
If $s=50,$ then
$h={\sqrt {50^{2}-30^{2}}}=40.$
So, we have
$2(40)6=2(50)s'.$
Solving for $s',$ we get $s'={\frac {24}{5}}{\text{ m/s.}}$

Final Answer:
$s'={\frac {24}{5}}{\text{ m/s}}$

@@ Line 33: / Line 33: @@
 !Step 2: &nbsp;
 |-
-|If &nbsp; <math style="vertical-align: -4px">s=50,</math>&nbsp; then &nbsp;<math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
+|If &nbsp; <math style="vertical-align: -4px">s=50,</math>&nbsp; then
 |-
-|So, we have &nbsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
 |-
-|Solving for &nbsp; <math style="vertical-align: -5px">s',</math>&nbsp;  we get &nbsp; <math style="vertical-align: -14px">s'=\frac{24}{5}</math> &nbsp; m/s.
+|So, we have
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
+|-
+|Solving for &nbsp; <math style="vertical-align: -5px">s',</math>&nbsp;  we get &nbsp; <math style="vertical-align: -14px">s'=\frac{24}{5} \text{ m/s.}</math> &nbsp;
 |}
@@ Line 44: / Line 48: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">s'=\frac{24}{5}</math>&nbsp; m/s
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">s'=\frac{24}{5} \text{ m/s}</math>&nbsp;
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]