Related Rates - Differential Equations

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Homework Help Overview

The problem involves a ladder of length 13m leaning against a wall, with the base sliding away from the wall. Participants are tasked with finding the rate at which the top of the ladder slides down the wall and the rate of change of the angle subtended by the ladder with the ground.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply implicit differentiation and the chain rule to relate the rates of change of the ladder's position and angle. Some participants suggest differentiating with respect to time instead of distance to find the angle's rate of change.

Discussion Status

Participants are exploring different methods of differentiation, particularly focusing on the use of the chain rule and implicit differentiation. There is acknowledgment of the need to follow specific scoring criteria for the exercise, which influences the approach taken.

Contextual Notes

Participants mention constraints related to scoring points for using certain mathematical techniques, such as implicit differentiation and the chain rule, which may affect their problem-solving strategies.

Cilabitaon
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Homework Statement


A ladder, of length 13m, is leaning against a wall. The base of the ladder begins to slide, frictionless, away from the wall; by the time the base is horizontally 12m away from the wall it is moving at a rate of 2ms-1

a)How fast is the top of the ladder sliding down the wall?

b)At what rate is the angle subtended from the ground by the ladder changing by?

Homework Equations


Assuming that
\dot{y} = \frac{dy}{dt},
\dot{x} = \frac{dx}{dt},
y' = \frac{dy}{dx}

y' = \dot{y} \times \dot{x}^{-1} (chain rule)
Implicit Differentiation(basic)
Pythagoras' Theorem

The Attempt at a Solution


First I drew a right angle triangle with the ladder as the hypotenuse and labelled the sides y(the wall) and x(the ground).

Then I get that, when;
x = 12 , \dot{x} = 2

Also;
y = \sqrt{13^{2} - x^{2}}

The next part I did only because I had to use implicit differentiation at some point in this.

y^{2} = 13^{2} - x^{2}

2yy' = -2x

y' = \frac{-2x}{2y}

Now rearrange the chain rule equation to get;

\dot{y} = y' \times \dot{x}

\dot{y} = \frac{-4x}{2y}

\dot{y} = \frac{-2x}{y}

To which I then sub in my values for when x = 12

\dot{y} = \frac{-2(12)}{({13^{2} - 12^{2})^{\frac{1}{2}}}

\dot{y} = -4.8ms^{-1}

I think the first part is right, but I am having trouble with part (b)

I assume it has something to do with
\tan\Theta = \frac{y}{x}

But I can't differentiate that(actually, I can, I just can't seem to get any sensible kind of answer).

All I'm thinking right now is;
\frac{d}{dx}(\tan\Theta) = -\frac{y}{x^2}

Does anybody know if this is the right way to go?
 
Last edited:
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Since you're looking for dθ/dt, try differentiating with respect to t instead of x.
 
vela said:
Since you're looking for dθ/dt, try differentiating with respect to t instead of x.

I would, but the point is I have to use the chain rule as many times as I can. The points scored in the exercise are 5 for a correct answer and 2 for either the use of implicit differentiation or the chain rule. This is basic, I know, but we fail if we get less than 16 points.
 
You get to use the chain rule and implicit differentiation if you differentiate with respect to t.
 

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