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This is another Khan Related Rates problem that I haven't been able to understand where I'm going wrong.

GIVEN:

A 13 meter ladder is leaning against a wall when its base begins to slide away. By the time the base is 12 meters from the wall, it is moving at a rate of dx/dt = 5 m/sec. At that moment, at what rate is the angle dθ/dt between the ground and the ladder changing (in radians/sec)?

Your answer should be negative if the angle is getting smaller.

/GIVEN

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I decided to approach this using tan(θ).

MY STEPS:

tan(θ) = y/x

θ = atan(y/12)

d/dt(θ) = d/dt(atan(y/12))

d/dt(θ) = 1/(1 + (y/12)^2)*d/dt(y/12)

d/dt(θ) = 1/(12*(1 + (y/12)^2))*d/dt(y)

d/dt(θ) = 1/(12*(1 + (y/12)^2))*dy/dt

d/dt(θ) = 1/(12*(1 + y^2/12^2))*dy/dt

d/dt(θ) = 1/(12*(1 + y^2/144))*dy/dt

d/dt(θ) = 1/(12 + 12*y^2/144)*dy/dt

d/dt(θ) = 1/(144/12 + y^2/12)*dy/dt

d/dt(θ) = 1/(144 + y^2)/12*dy/dt

dθ/dt = [12/(y^2 + 144)*dy/dt]

Finding y:

y = sqrt(13^2 - 12^2)

y = sqrt(169 - 144)

y = sqrt(25)

y = 5

Insert y:

dθ/dt = 12/(5^2 + 144)*dy/dt

dθ/dt = 12/(25 + 144)*dy/dt

dθ/dt = [12/169*dy/dt]

Find dy/dt:

y = sqrt(13^2 - x^2)

y = (13^2 - x^2)^1/2

d/dt(y) = d/dt((13^2 - x^2)^1/2)

d/dt(y) = 1/2*(13^2 - x^2)^-1/2*d/dt(13^2 - x^2)

d/dt(y) = 1/2*sqrt(13^2 - x^2)*d/dt(13^2 - x^2)

d/dt(y) = 1/2*sqrt(13^2 - x^2)*(d/dt(13^2) - d/dt(x^2))

d/dt(y) = 1/2*sqrt(13^2 - x^2)(d/dt(13^2) - 2xd/dt(x))

d/dt(y) = 1/2*sqrt(13^2 - x^2)(d/dt(13^2) - 2xdx/dt)

d/dt(y) = 1/2*sqrt(13^2 - x^2)(0 - 2xdx/dt)

d/dt(y) = 1/2*sqrt(13^2 - x^2)(-2xdx/dt)

d/dt(y) = -2x/2*sqrt(13^2 - x^2)*dx/dt

dy/dt = [-x/sqrt(13^2 - x^2)*dx/dt]

Insert x:

x = 12

dy/dt = -12/sqrt(13^2 - 12^2)*dx/dt

dy/dt = -12/sqrt(169 - 144)*dx/dt

dy/dt = -12/sqrt(25)*dx/dt

dy/dt = [-12/5*dx/dt]

Insert dx/dt:

dx/dt = 5

dy/dt = -12/5*5

dy/dt = -12*5/5

dy/dt = [-12]

Insert dy/dt:

dθ/dt = 12/169*-12

dθ/dt = 12*-12/169

dθ/dt = -144/169

dθ/dt ≈ [-0.85207100591715976331360946745562130177514792899408]

/MY STEPS

GIVEN SOLUTION:

Let x be the distance from the wall to the base of the ladder. We know that

cos(θ) = x/(ladder length) = x/13

When x = 12, we also know from the Pythagorean Theorem that if y is the height of the top of the ladder then

12^2 + y^2 = 13^2 or y^2 = 25 or y = 5

We can use this to determine that, when x = 12,

sin(θ) = 5/13

Let's now implicitly take the derivative of both sides of the equation

cos(θ) = x/13

with respect to time "t".

d/dt(cos(θ)) = d/dt(x/13)

(−sin(θ))dθ/dt = 1/13*dx/dt

dθ/dt = (1/(13*(−sin(θ))))*dx/dt

Then when x = 12, dx/dt = 5, and sin(θ) = 5/13:

dθ/dt = (5)(1/(13(-5/13)))

dθ/dt = (5)(1/(-5*13/13))

dθ/dt = (5)(1/(-5))

dθ/dt = (5)(-1/5)

dθ/dt = -5/5

dθ/dt = −1 radian/sec.

/GIVEN SOLUTION

What could cause the significant discrepancy.

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MY STEPS:

tan(θ) = y/x

θ = atan(y/12)

d/dt(θ) = d/dt(atan(y/12))

The problem here is that $x$ isn't constant. So, you should first have $\theta = \arctan\dfrac{y}{x}$ and then differentiate to get $\dfrac{d\theta}{dt} = \dfrac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \dfrac{x\frac{dy}{dx}-y\frac{dx}{dt}}{x^2}$.

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  • $\begingroup$ Not sure what I was thinking by processing x as a constant. Lesson learned. Thanks! $\endgroup$ – redthumb Aug 12 '14 at 10:50

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