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

Revision as of 15:11, 22 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 $30^{2}+h^{2}=s^{2}$ by the Pythagorean Theorem.
Taking derivatives, we get
$2hh'=2ss'$ .

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

@@ Line 16: / Line 16: @@
 |Insert diagram.
 |-
-|From the diagram, we have <math>30^2+h^2=s^2</math> by the Pythagorean Theorem.
+|From the diagram, we have <math style="vertical-align: -2px">30^2+h^2=s^2</math> by the Pythagorean Theorem.
 |-
 |Taking derivatives, we get
 |-
-|<math>2hh'=2ss'</math>.
+|
+::<math>2hh'=2ss'</math>.
 |}
@@ Line 26: / Line 27: @@
 !Step 2: &nbsp;
 |-
-|If <math>s=50</math>, then <math>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>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>s'</math>, we get <math>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.
 |}