Pg. 223 # 17 A 13-ft ladder is leaning against a wall. If the top of the ladder slips down the wall at a rate of 2 ft/s, h...
Step 1 <ul><li>  Draw a diagram + What do we know?  </li></ul>y  13 x Ladder: 13 ft dy/dt= 2 ft/s dx/dt= ?  When y=5 ft
Step 2 <ul><li>Use the Pythagorean theorem to establish  </li></ul><ul><li>a relationship between the variables: </li></ul...
Step 3 <ul><li>Take the derivative of the equation: </li></ul><ul><li>2x(dx/dt) + 2y(dy/dt)= 0  </li></ul><ul><li>  Don’t ...
Step 4 <ul><li>  Find x when y=5 using the  Pythagorean  theorem: </li></ul>13 x 5   (x)^2 + (5)^2=(13)^2   (x)^2 + 25=169...
Step 5 <ul><li>Now plug in what you know… </li></ul><ul><li>  2x(dx/dt) + 2y(dy/dt)= 0  </li></ul><ul><li>  2(12)(dx/dt) +...
Step 6 <ul><li>… and solve for dx/dt! </li></ul><ul><li>2(12)(dx/dt) + (2)(5)(2)= 0 </li></ul><ul><li>24 (dx/dt) + (20)= 0...
Step 7 <ul><li>Be sure to answer the question! </li></ul><ul><li>Therefore, when the top is 5 ft above the  ground, the fo...
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  1. 1. Pg. 223 # 17 A 13-ft ladder is leaning against a wall. If the top of the ladder slips down the wall at a rate of 2 ft/s, how fast will the foot be moving away from the wall when the top is 5 ft above the ground?
  2. 2. Step 1 <ul><li> Draw a diagram + What do we know? </li></ul>y 13 x Ladder: 13 ft dy/dt= 2 ft/s dx/dt= ? When y=5 ft
  3. 3. Step 2 <ul><li>Use the Pythagorean theorem to establish </li></ul><ul><li>a relationship between the variables: </li></ul><ul><li> (x)^2 + (y)^2=(13)^2 </li></ul>
  4. 4. Step 3 <ul><li>Take the derivative of the equation: </li></ul><ul><li>2x(dx/dt) + 2y(dy/dt)= 0 </li></ul><ul><li> Don’t forget: The derivative of a constant is zero! </li></ul>
  5. 5. Step 4 <ul><li> Find x when y=5 using the Pythagorean theorem: </li></ul>13 x 5 (x)^2 + (5)^2=(13)^2 (x)^2 + 25=169 (x)^2 =144 Now: take the positive square root of both sides x= 12
  6. 6. Step 5 <ul><li>Now plug in what you know… </li></ul><ul><li> 2x(dx/dt) + 2y(dy/dt)= 0 </li></ul><ul><li> 2(12)(dx/dt) + (2)(5)(2)= 0 </li></ul>
  7. 7. Step 6 <ul><li>… and solve for dx/dt! </li></ul><ul><li>2(12)(dx/dt) + (2)(5)(2)= 0 </li></ul><ul><li>24 (dx/dt) + (20)= 0 </li></ul><ul><li>24 (dx/dt)= -20 </li></ul><ul><li> Now divide both sides by 24 </li></ul><ul><li> dx/dt= -24/20 or -5/6 </li></ul>
  8. 8. Step 7 <ul><li>Be sure to answer the question! </li></ul><ul><li>Therefore, when the top is 5 ft above the ground, the foot will be moving away from wall at the rate of (5/6) ft/sec </li></ul>

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