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| # h3cadiap2 → rates of change |
|
A 26 foot ladder is placed against a wall (Fig.1). If the top of the ladder is sliding down the wall at 2 feet per second, at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall? |
(Problem credited by -- High School level)
 |
|
| The rate of of the bottom of the ladder moving away from the wall → |
dx
dt
|
| Also take diff both side from the equation(1) to find → |
dx
dt
|
| Solution |
| The ladder is a constant length, the bottom of the ladder will move away from the wall at the same rate that the top of the ladder is moving down the wall. At any moment in time, let x be the distance of the bottom of the laddfer from the wall, and let Y be the distance of the top of the ladder on the wall (see Figure 1). Both x and y are changing with respect to time and can be thought of as functions of time ; that is , x=x(t) and y=y(t). Furthermore, x and y are related by ther Pythagorean relationship: |
| x2 + y2 = 262 → (2) |
| Take Diff both side; |
dx
dt
|
2x | + |
dy
dt
|
2y | = | 0 | → (3) |
| Problem gives ; |
dy
dt
|
=-2 (y is decreasing at a constant rate of 2 feet per second), |
| problem is to find |
dx
dt
|
when x =10 feet |
| 102 + y2 | = | 262 |
| y | = | 24 feet |
 | ||
| Substitute |
dy
dt
|
= -2, x = 10, and y =24 into (2), then solve for |
dx
dt
|
| 2(10) |
dx
dt
|
+ 2(24)(-2) = 0 |
| ∴ |
dx
dt
|
= | 4.8 feet·sec-1 |
|
Ans.
4.8 feet·sec-1
is the rate of the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall |
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