The top of a ladder slides down a vertical wall at a rate of 0.15 m/s. At the moment when the bottom of the ladder is 3m from the wall, it slides away from the wall at a rate of 0.2 m/s. How long is the ladder?
I understand what the question is saying and I kind of get how to set up the equation but I'm getting lost on how to solve it. Please answer this with steps. Thanks.
I understand what the question is saying and I kind of get how to set up the equation but I'm getting lost on how to solve it. Please answer this with steps. Thanks.
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Pythagorean theorem
a^2 +b^2 = c^2
``|\
a|. \c
..|__\
b
the c drops out because its rate is 0
now your formula is:
a^2 = b^2
do your implicit differentiation and solve with what you know
2a*a' = 2b*b'
2a*(0.15) = 2*(3)*(0.2)
a = 4
now back to the original formula...
(4)^2 +(3)^2 = C^2
5 = C = height of ladder
a^2 +b^2 = c^2
``|\
a|. \c
..|__\
b
the c drops out because its rate is 0
now your formula is:
a^2 = b^2
do your implicit differentiation and solve with what you know
2a*a' = 2b*b'
2a*(0.15) = 2*(3)*(0.2)
a = 4
now back to the original formula...
(4)^2 +(3)^2 = C^2
5 = C = height of ladder
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Call the vertical side "y", and the horizontal side "x". By Pythagoras...
x² + y² = h²
Differentiating through the chain rule results in:
2x(dx/dt) + 2y(dy/dt) = 0
Notice that it's equated to 0. This is because the hypotenuse does not change in length, and as such, it's derivative is zero. If you took a snapshop of the ladder after a certain amount of time, the height against the wall and the length along the ground will be different, but the length of the ladder stays consistent.
So now, plug and chug.
2*3*(0.2) + 2y(-0.15) = 0
(Note that I made 0.15 negative because it's sliding downwards).
1.2 = 0.3y
y = 4
So when x = 3, y = 4. Either recognize the Pythagorean triple, or use the equation:
h = √(3² + 4²) = √25 = 5 m
The ladder is 5 meters long.
Hope this helps.
x² + y² = h²
Differentiating through the chain rule results in:
2x(dx/dt) + 2y(dy/dt) = 0
Notice that it's equated to 0. This is because the hypotenuse does not change in length, and as such, it's derivative is zero. If you took a snapshop of the ladder after a certain amount of time, the height against the wall and the length along the ground will be different, but the length of the ladder stays consistent.
So now, plug and chug.
2*3*(0.2) + 2y(-0.15) = 0
(Note that I made 0.15 negative because it's sliding downwards).
1.2 = 0.3y
y = 4
So when x = 3, y = 4. Either recognize the Pythagorean triple, or use the equation:
h = √(3² + 4²) = √25 = 5 m
The ladder is 5 meters long.
Hope this helps.