APPLICATIONS OF DERIVATIVES
The top of a ladder 13m long is resting against a vertical wall when a ladder begins to slide. When the foot of the ladder is 5m from the wall, it is sliding at the rate of 2m/s. How fast then is the top sliding downwards?
Answers
Step-by-step explanation:
A 13m long ladder is leaning against a wall. The bottom of the ladder is pulled away from the wall, along the ground, at the rate of 2m/s. How fast is the height of the wall decreasing when the foot of the ladder is 5m away from the wall?
Answer is (5/6) m/s
The ladder will form the hypotenuse. Length of hypotenuse = 13m.
Let us take the horizontal distance between the wall and the ladder as ‘ground’. This g = 5m
wall ^ 2 + ground ^ 2 = 13^2 = 169
w^2 + g^2 = 169
w^2 + 25 = 169. Thus, w^2 = 144. Thus, w = 12
Rate of change of ground distance = dg / dt = 2m/s (given).
In the equation w^2 + g^2 = 169, differentiate the equation with respect to time.
2 w dw/dt + 2 g dg/dt = 0
2 (12) (dw/dt) + 2 (5) (2) = 0. So, 24 dw/dt + 20 = 0
dw/dt = -20 / 24 = -5 / 6
Thus, the height of the wall is decreasing at a rate of -5/6 m / s (that is, -0.833m/s)
The minus sign denotes that the height of the ladder is falling.
Answer:
Refer the above attachment!
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