Find the velocity of the shadow variation

[adriaat]

Helo,
I've been working out this exercise, but my solution and the text's aren't the same.

Homework Statement
We have a spotlight on the floor located at a distance of 30.5 m from a wall of a building. There is a person 1.83 m tall between the spotlight and the wall moving away from the spotlight at a constant velocity of 1.83 m/s. Calculate the velocity of the shadow variation when the person is at 9.14 m from the spotlight.

So, let's name
$x_{1}=9.14 m$
$x_{2}=30.5 m$
$y_{1}=$the height of the shadow when the person is on $x_{1}$
$y_{2}=$the height of the shadow when the person is on $x_{2}$
$h=1.83 m$ (height of the person)
$v=1.83 m/s$ (velocity)

The attempt to a solution

1) I find the height of the shadow projected on the wall when the body is on x₁. To do this, I consider the spotlight and the person make a right triangle, with angle $\alpha$ which is the same of the triangle made with the height of the shadow y₁ and the distance between the spotlight and the wall.

$\alpha=arctan(\frac{h}{x_{1}})$

And now I find $y_{1}$

$y_{1}=tan \alpha·x_{2}=6.11 m$

Then the height of the shadow at $x_{2}$ will be $y_{2}=1.83 m$. I think this is obvious.

Afterwards, I want to find the time that took place moving from $x_{1}$ to $x_{2}$ because this will be the same time from $y_{1}$ to $y_{2}$

$\Delta t=\frac{\Delta x}{v}=11.67 s$

Then, to conclude

$v_{shadow}=\frac{\Delta y}{\Delta t}=-0.366 m/s$

The solution of the book is
$v_{shadow}=-1.22 m/s$

Anyone can help me, please?

 

[collinsmark]

Where does the $x_1 = 9.14 \ \mathrm{m}$ come from? I see it written down in your variable name definitions, but it's not in the original problem statement.

Is this a calculus based class? The easiest way (at least for me) to approach this problem involves differentiation. Maybe we can solve this problem by other means, but if you are allowed to use differential calculus, I would advise using it.

---------------------

You can combine your

$\alpha = \arctan \frac{h}{x_1}$

and

$y_1 = x_2 \tan \alpha$

into one simple equation that doesn't involve any trigonometric functions. That will make the coming math quite a bit easier.

After that, don't plug any numbers in just yet. Now is the time for differentiation.

 

[adriaat]

Thank you, I didn't write that [...]calculate the velocity of the shadow variation when the person is at 9.14 m from the spotlight.

Yes, it is a calculus based class, but I just started calculus classes, so I'm not familiar with differentiation.
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To combine the equations I do:
$\arctan \frac{y_1}{x_2} = \alpha$
$\alpha = \arctan \frac{h}{x_1}$

Then
$\frac{y_2}{x_2} = \frac{h}{x_1}$

I get stuck here. Do not know sure how to differentiate this equation or if it's possible.

 

[collinsmark]

Thank you, I didn't write that [...]calculate the velocity of the shadow variation when the person is at 9.14 m from the spotlight.

That makes sense.

Yes, it is a calculus based class, but I just started calculus classes, so I'm not familiar with differentiation.

I can't think of any other way to solve the problem at the moment. I'm afraid you're going to have to take a derivative, one way or the other.

To combine the equations I do:
$\arctan \frac{y_1}{x_2} = \alpha$
$\alpha = \arctan \frac{h}{x_1}$

Then
$\frac{y_2}{x_2} = \frac{h}{x_1}$

I get stuck here. Do not know sure how to differentiate this equation or if it's possible.

So far so good.

But allow me to rearrange some variables:

$$y_2 = \frac{x_2 h}{x_1}$$
Note that $x_2$ and $h$ are constants. They don't change with time. On the other hand, $x_1$ is not constant. $x_1$ is a function of time. Also, $y_2$ is a function of $x_1$ making $y_2$ also a function of time.

Ultimately what you are looking for is $\frac{dy_2}{dt},$ the velocity of the shadow.

You have a choice of different approaches at this point.

You could take the derivative of the equation with respect to time, paying special attention to the fact that $x_1$ is itself a function of time.

Or you could take an intermediate step and find $\frac{dy_2}{dx_1},$ and then find a relationship between $\frac{dy_2}{dx_1}$ and $\frac{dy_2}{dt}$ as a second step.

[Edit: minor rewording.]

 

[adriaat]

I found a solution which agrees with my text's.

Here is what I did:

1st equation: $\frac{dy_2}{dt}=x_2·h·\frac{1}{x_1}=x_2·h·\frac{1}{v·t}$

I derivated the above equation, which is the velocity of the shadow's variation, and found time for $x_1$

2nd equation: $t=\frac{x_1}{v}$

I combined the 2nd equation into the first one derivated, calculated and as a result I got $-1.22 \text m\text / \text s$

I think it's OK,

Thank you a lot for your dedication (:

 

[collinsmark]

I found a solution which agrees with my text's.

Your approach isn't the approach I would have taken, but it's still valid. However, there are some things that don't seem quite right, as I'll describe below.

Here is what I did:

1st equation: $\frac{dy_2}{dt}=x_2·h·\frac{1}{x_1}=x_2·h·\frac{1}{v·t}$

Although it's not necessary, it is okay to make the $vt = x_1$ substitution, since the person is moving at a constant velocity. It also assumes that $x_1 = 0$ at time $t = 0$, which is fine for the purposes of this problem.

The problem I see is that the right hand side of your above equation is before the derivative is taken, not after.

$$y_2 = x_2 h \left( \frac{1}{vt} \right) = \frac{x_2 h}{v} \left(\frac{1}{t} \right)$$

But that's the relationship before the derivative, not after. Now take the derivative,

$$\frac{dy_2}{dt} = \frac{d}{dt} \left\{\frac{x_2 h}{v} \left(\frac{1}{t} \right) \right\} = \frac{x_2 h}{v} \frac{d}{dt} \left\{ \frac{1}{t} \right\}$$

I derivated the above equation, which is the velocity of the shadow's variation, and found time for $x_1$

2nd equation: $t=\frac{x_1}{v}$

I combined the 2nd equation into the first one derivated, calculated and as a result I got $-1.22 \text m\text / \text s$

I think it's OK,

Thank you a lot for your dedication (:

I'm glad you got the answer you were looking for, but I'm just hoping that you didn't stumble upon the right answer for the wrong reasons, by sheer coincidence.

Personally, I solved the problem by differentiating with respect to $x_1$, obtaining an expression for $\frac{dy_2}{dx_1}$. Then realizing $\frac{dy_2}{dx_1} = \frac{dy_2}{dt} \frac{dt}{dx_1}$, also realizing that $v = \frac{dx_1}{dt}$, substituting, and finally solving for $\frac{dy_2}{dt}$.