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

Revision as of 12:32, 29 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:
Recall:
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
$2hh'=2ss'.$

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

@@ Line 11: / Line 11: @@
 |-
 |
-::hypotenuse, we have <math style="vertical-align: -2px">a^2+b^2=c^2</math>.
+::hypotenuse, we have <math style="vertical-align: -2px">a^2+b^2=c^2.</math>
 |}
@@ Line 26: / Line 26: @@
 |-
 |
-::<math>2hh'=2ss'</math>.
+::<math>2hh'=2ss'.</math>
 |}
@@ Line 32: / Line 32: @@
 !Step 2: &nbsp;
 |-
-|If <math style="vertical-align: -1px">s=50</math>, then <math style="vertical-align: -3px">h=\sqrt{50^2-30^2}=40</math>.
+|If <math style="vertical-align: -1px">s=50,</math> then <math style="vertical-align: -3px">h=\sqrt{50^2-30^2}=40.</math>
 |-
-|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.
 |}