Author Topic: Related rates of change (Read 2356 times) Tweet Share

dinosaur93 · « **on:** August 04, 2012, 09:43:33 pm »

How do you do this type of problems where you find to find more than 1 unknown? I know that we need to find $\frac {dy}{dt}$ in get to $\frac {dA}{dt}$

A 13 feet long ladder rests against a vertical wall. If the foot of the ladder is pulled away from the wall at a constant rate of 8 inches per seconds, find the rate at which the area of the triangle is formed by the ladder when the wall and the ground changing at the instant when the top of the ladder is 12 feet above the ground.

From the question, I know that the unknown is $\frac {dA}{dt}$ when y=12 ft
I also know that $\frac {dx}{dt} = \frac {2}{3} ft/sec$
from memory = $A = \frac{1}{2}bh$

How do we find $\frac {dy}{dt}$ ? Do we just use the pythagoras theorem $x^2 + y^2 = 13ft$ put everything in terms of x, then differentiate it??? then multiply by $\frac {dx}{dt}$ to get $\frac {dy}{dt}$

? The final answer gives $\frac {119}{36} ft^2/sec$ which Im no where closed to...

kamil9876 · « **Reply #1 on:** August 04, 2012, 11:43:07 pm »

Let x be the distance between the bottom of the wall and the bottom of the ladder. Let y be the distance from bottom of wall to top of ladder. Then $\frac{dx}{dt}=8$ .

$A=xy/2$

One approach is to express in terms of x only and then differentiate to find $\frac{dA}{dx}$ , then use $\frac{dA}{dt}=\frac{dA}{dx}\frac{dx}{dt}$

Another way is to differentiate both sides with respect to t and use the product rule to get $\frac{dA}{dt}=\frac{1}{2}(\frac{dx}{dt}y+\frac{dy}{dt}x)$ . Then use dy/dt=dy/dx dx/dt

Phy124 · « **Reply #2 on:** August 04, 2012, 11:47:37 pm »

As you said we have $\frac{dx}{dt}$ $(\frac{2}{3}ft/sec)$ and we want $\frac{dA}{dt}$

So we can use $\frac{dA}{dt}=\frac{dA}{dx}\frac{dx}{dt}$ .

So we need to find $\frac{dA}{dx}$

The area of a triangle is given by the formula $A=\frac{1}{2}bh$ or in this case $A=\frac{1}{2}xy$

But we only want A in terms of x so that we can differentiate A with respect to x. To do this we have to find y in terms of x.

We can do this using pythagoras' theorem;

$x^2+y^2=13^2$

$\therefore y =\sqrt{13^2-x^2} = \sqrt{169-x^2}$

Thus;

$A=\frac{1}{2}x\sqrt{169-x^2}$

and hence using the product rule;

$\frac{dA}{dt}=\frac{169-2x^2}{2\sqrt{169-x^2}}$

So at the moment we have;

$\frac{dA}{dt}=\frac{dA}{dx}\frac{dx}{dt} = \frac{2}{3} \times \frac{169-2x^2}{2\sqrt{169-x^2}}$

We want to find $\frac{dA}{dt}$ when the height is 12 feet (when y = 12) but our equation is in terms of x, so we must find x.

$x=\sqrt{169-12^2}=\sqrt{169-144}=\sqrt{25}=5$

$\frac{dA}{dt}=\frac{2}{3} \times \frac{169-2(5)^2}{2\sqrt{169-(5)^2}}=\frac{119}{36}ft^2/sec$

dinosaur93 · « **Reply #3 on:** August 05, 2012, 05:06:56 pm »

awesome!!! thank you very much!!!~ your method is very easy to understand and straight forward

another Qs, my friend derived with the same answer as you but in a different method
nad I dont seem to understand how he did some of the steps? could someone help me explain thank you!!

How did he derived from $\frac{d}{dt} (A) = \frac{d}{dt} (\frac{1}{2}xy) \Leftrightarrow$ to $\frac{dA}{dt} = \frac{1}{2}x\frac{dy}{dt} + \frac{1}{2}y\frac{dx}{dt}$

$x^{2} + y^{2} = 13^{2}$ to $\Leftrightarrow 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$= \frac{dy}{dt} = \frac{-x}{y}\frac{dx}{dt}$

$\frac{dy}{dt}\left \right |_{y = 12ft} = \frac{-5}{12} \times \frac{2}{3}= \frac{-5}{18}ft/sec$

So now we can find $\frac{dA}{dt}\left \right |_{y = 12ft}$

$\frac{dA}{dt} \left \right |_{y = 12ft}= (\frac{1}{2}x\frac{dy}{dt} + \frac{1}{2}y\frac{dx}{dt})\left \right |_{y = 12ft}$

$\frac{1}{2}(5ft)(\frac{-5}{18}ft/sec) + \frac{1}{2}(12ft)(\frac{2}{3}ft/sec)$

$=\frac{25}{36}ft^2/sec + \frac{144}{36}ft^2/sec$

$= \mathbf{\frac{119}{36}ft^2/sec}$

Jenny_2108 · « **Reply #4 on:** August 05, 2012, 05:17:24 pm »

The way your friend did is in Spesh course, not in Methods.
Its differentiation of equation. I reckon you do as rangaa because its safer. I heard from 2011 students, VCAA may not accept Spesh knowledge applying in Methods

dinosaur93 · « **Reply #5 on:** August 05, 2012, 05:45:26 pm »

oh, i see....

no wonder the solution seems a bit odd....

thank you for clarifying!~

dinosaur93 · « **Reply #6 on:** August 05, 2012, 05:59:02 pm »

How about if we were to find the the $\frac {d \theta}{dt}$ using the Methods way
I know how to do it by differentiating the equation....but since its not preferred in the methods course, I dont think I can obtain full marks in the working out section even though you get the right answer....

A 13 feet long ladder rests against a vertical wall. If the foot of the ladder is pulled away from the wall at a constant rate of 2 inches per seconds, how fast is the triangle is formed by the ladder ~~when the wall~~ and the ground changing (in radians per seconds) at the instant when the top of the ladder is 12 feet above the ground?

What I did is I differentiate the equation $\mathbf{\cos (\theta) = \frac {x}{13}}$ then re-arrange the equation to find $\mathbf{\frac {d \theta}{dt}}$ and subsituting $\mathbf{\frac {dx}{dt}}$ and y = 12 into

the trig expression $\mathbf {-\sin (\theta)}$ to get an answer of $\mathbf{- \frac {1}{72} radians/sec}$

EDIT: the question is ask for radians per seconds

Jenny_2108 · « **Reply #7 on:** August 05, 2012, 06:18:49 pm »

Quote from: elvin.lam1 on August 05, 2012, 05:59:02 pm

What I did is I differentiate the equation $\mathbf{\cos (\theta) = \frac {x}{13}}$ then re-arrange the equation to find $\mathbf{\frac {d \theta}{dt}}$ and subsituting $\mathbf{\frac {dx}{dt}}$ and y = 12 into

the trig expression $\mathbf {-\sin (\theta)}$ to get an answer of $\mathbf{- \frac {1}{72} radians/sec}$

But the question asks about $\mathbf{\frac {dA}{dt}}$ , not $\mathbf{\frac {d \theta}{dt}}$

dinosaur93 · « **Reply #8 on:** August 05, 2012, 06:24:12 pm »

Quote from: Jenny_2108 on August 05, 2012, 06:18:49 pm

But the question asks about $\mathbf{\frac {dA}{dt}}$ , not $\mathbf{\frac {d \theta}{dt}}$

But the question is asking for radians per seconds though, not the area...isnt it?

sorry for the confusion...

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