How Is the Rate of Approach to a Tower's Top Calculated Using Derivatives?

[adjacent]

Homework Statement

A man is walking at the rate of 6.5km/h towards the foot of a tower 120m high.At what rate is he approaching the top of the tower when he is 50m away from the bottom of the tower?

Homework Equations

Nah

The Attempt at a Solution

$6.5km/h = 1\frac{29}{36}m/s$
I know how to find the derivative of many functions but I don't know how to apply it here.Any hint?

*Note: I haven't officially taken a calculus course. I am self learning it. So please bear with me

 

[hilbert2]

Let's call the distance from the top of the tower $s_{t}$ and the distance from the bottom $s_{b}$. Now, by the Pythagorean theorem, we have $s_{t}=\sqrt{(120m)^2+s^{2}_{b}}$. Differentiating both sides with respect to time $t$, we have

$\frac{ds_{t}}{dt}=\frac{d}{dt}\sqrt{(120m)^2+s^{2}_{b}}$. Note that $\frac{ds_{t}}{dt}$ is the speed with which the man approaches the top and $\frac{ds_{b}}{dt}$ is the speed with which he approaches the bottom. All you need to do is calculate the time derivative of the square root expression and substitute the values you have been given.

 

[adjacent]

All you need to do is calculate the time derivative of the square root expression and substitute the values you have been given.

$\frac{d}{dt}\left(\sqrt{120^2+s_{b}^2}\right)=\frac{1}{2}(120^2+s_{b}^2)^{\frac{-1}{2}}.2s_b$?

 

[LCKurtz]

$\frac{d}{dt}\left(\sqrt{120^2+s_{b}^2}\right)=\frac{1}{2}(120^2+s_{b}^2)^{\frac{-1}{2}}\cdot 2s_b$?

Not quite. What you have calculated is $\frac{d}{ds_b}\left(\sqrt{120^2+s_{b}^2}\right)$, not $\frac{d}{dt}$. You need the chain rule.

 

[BiGyElLoWhAt]

Are you familiar with the chain rule for derivatives? It might be easier to use that here, given the expression you have.

Spoiler

$\frac{df}{dt} = \frac{df}{ds_b}\frac{ds_b}{dt}$

 

[BiGyElLoWhAt]

You need the chain rule.

OOps didn't see that LC. Not trying to steal your thunder XD

 

[adjacent]

Not quite. What you have calculated is $\frac{d}{ds_b}\left(\sqrt{120^2+s_{b}^2}\right)$, not $\frac{d}{dt}$. You need the chain rule.

From where does this $t$ come from? There are no t s here

Yes I know chain rule.
$(f(g(x)))'=f'(g(x)).g'(x)$
I don't understand that fraction form very well

 

[CAF123]

From where does this $t$ come from? There are not t s here

The rate of approach of the man is equivalent to the rate of change of his displacement. This is (d/dt) s

Yes I know chain rule.
$(f(g(x)))'=f'(g(x)).g'(x)$
I don't understand that fraction form very well

In this case, the distance of the man from the bottom of the tower is changing as t increases. So you have that s_b = s_b(t). But you know that the distance of the man from the top is a function of s_b. So s_t = s_t(s_b) = s_t(s_b(t)). f'(g(x)) means the derivative of f wrt the argument g(x). See if you can identify f(g(x)) and g(x) in your problem.

 

[adjacent]

This is confusing. Let me start from the OP.
Let the distance from the top of the tower be $s_1$ and let the distance from the bottom of the tower be $s_2$.
We know that $s_1=\sqrt{120^2+s_{2}^2}$
Then what do I have to do? What is this time derivative? It's still unclear to me.

 

[hilbert2]

If at time t=0 the man is 50 m away from the bottom of the tower and he is approaching the bottom with speed 6.5 km/h, then obviously the distance $s_2$ is a function of time: $s_2 = 0.05km-(6.5\frac{km}{h})\times t$. Obviously, $s_1$ is also a function of t. This has just not been explicitly written in the problem statement.

 

[CAF123]

This is confusing. Let me start from the OP.
Let the distance from the top of the tower be $s_1$ and let the distance from the bottom of the tower be $s_2$.
We know that $s_1=\sqrt{120^2+s_{2}^2}$
Then what do I have to do? What is this time derivative? It's still unclear to me.

You then want to find the rate of approach of the man when he is at some distance $s_2$ from the bottom of the tower. The rate he is approaching the top of the tower is given by the time derivative of his displacement from the top of the tower. So the next step would be to compute $\text{d}/\text{d}t\,\, s_1$. Does it make sense up to here?

Initially, you may think that since the expression for $s_1$ does not have any t's in it, that d/dt s₁ is simply zero. But it is not the case, because there is an implicit dependence of $s_2$ on t. That also makes sense right? I.e if the man keeps walking towards the tower, intuitively as time passes he gets closer to the base. So, in fact, $s_2 = s_2(t)$. The notation means that $s_2$ is a function of t, in the same way that f=f(x) means f is a function of x.

The chain rule allows you to extract the time dependence of $s_2$. To understand the notation both as you wrote and in 'fractions', it is that $$(s_1(s_2(t)))' = \frac{\text{d}s_1}{\text{d}t}\,\,\,\,;\,\,\,\, s_1'(s_2(t)) = \frac{\text{d}s_1}{\text{d}s_2}\,\,\,\,;\,\,\,\, s_2'(t) = \frac{\text{d}s_2}{\text{d}t} ,\,\,\,\,\,s_1 = s_1(s_2(t))$$

And again, $s_1 = s_1(s_2(t))$ meaning $s_1$ is a function of $s_2$, which is also a function of time.

 

[adjacent]

Oh. I see now. I think I will understand it better when I solve the problem. So:
Step 1:$s_1=\sqrt{0.12^2+6.5t}$
Is this correct?
*I may look like an idiot asking such question *

 

[Fredrik]

Oh. I see now. I think I will understand it better when I solve the problem. So:
Step 1:$s_1=\sqrt{0.12^2+6.5t}$
Is this correct?
*I may look like an idiot asking such question *

With your definitions of $s_1$ and $s_2$, we have $s_1=\sqrt{0.12^2+(s_2)^2}$. You should use your own definition of $s_2$ to write down a formula for $s_2$. (Hint: $s_2\neq\sqrt{6.5t}$).

If you prefer, you can write $s_1(t)$ and $s_2(t)$ instead of $s_1$ and $s_2$, to remind yourself of the time dependence. In your notation, $s_1$ and $s_2$ are variables whose values are determined by the value of another variable t. In the alternative notation, $s_1$ and $s_2$ are functions. This might make it easier to see how this is a job for the chain rule.

I also have to nitpick the notation $f(g(x))'=f'(g(x))g'(x)$ that you used earlier. (The left-hand side is very ugly). In an expression like f(x), f is the function, and f(x) is an element of its range. So f(x) is usually a number, not a function. You write f'(x) rather than f(x)' because you're taking the derivative of the function f, not the number f(x). The derivative of f is a function, and that function is denoted by f'. f'(x) denotes the value of that function at x. So the notation (f(g(x))' is very ugly to me. The best notation is $(f\circ g)'(x)$, but $\frac{d}{dx}f(g(x))$ is OK too.

The latter notation is kind of strange. The best way to interpret it is this: The x in the d/dx tells us two things: 1. It tells us that the function that we're taking the derivative of is the one that takes each real number x to f(g(x)) rather than, say, the one that takes each real number y to f(g(x)). (This would be a constant function). 2. When we have found the derivative of that function, we then find its value at x, rather than, say, its value at y.

 

[adjacent]

Thanks
$s_2=6.5t$
So $s_1=\sqrt{0.12^2+(6.5t)^2}$
Then what do I have to do?

 

[Fredrik]

That's not quite correct. You said that the guy is walking towards the tower, so the distances should be decreasing with time, not increasing.

When you have found the right formula, the next step is to use your knowledge about how the distance varies with t to compute the velocity.

Edit: I just saw your comment about not having studied calculus, so I guess I should add that if $x:\mathbb R\to\mathbb R$ is a function that takes the time coordinate to a particle's position at time t, then the velocity at time t is x'(t). (This is the definition of velocity).

 

[adjacent]

That's not quite correct. You said that the guy is walking towards the tower, so the distances should be decreasing with time, not increasing.

Then I will need an initial distance,which is not given in the question.

 

[CAF123]

Then I will need an initial distance,which is not given in the question.

Indeed, but you do not need to fully parametrise $s_2$. Try to use what has been said about the chain rule. From that, you will see you only need (ds₂/dt) which is given in the problem statement.

 

[Fredrik]

Then I will need an initial distance,which is not given in the question.

Right, so call that distance d. Then you compute the velocity $-\frac{ds_2}{dt}$ at an arbitrary time t. The result will of course contain d. Then you find the value of t that puts him at a distance of 50m from the base of the tower. The answer to this contains d as well. Then you find the value of $-\frac{ds_2}{dt}$ at that time. This result should not contain d. The d's must cancel out somehow.

Edit: CAF123 is right. The problem is easier if you don't substitute $d-6.5t$ for $s_2$ in the formula $s_1=\sqrt{0.12^2+(s_2)^2}$. Keep that $s_2$ around until you have computed $\frac{ds_1}{dt}$. The result contains both $s_2$ and $\frac{ds_2}{dt}$. d doesn't have any effect on the value of $\frac{ds_2}{dt}$ at any time t, and it doesn't have any effect on the value of $s_2$ at the specific time when the guy is 50m from the foot of the tower.

 

[LCKurtz]

Sorry I've been away from this thread for so long. If $x$ is the distance of the man from the base of the tower and $s$ is the distance from the top, you should have started with the original equation$$
x^2+120^2 =s^2$$and realize that $x$ and $s$ are functions of $t$. Then simply differentiate this equation with respect to $t$:$$
2xx' +0 = 2ss'$$This is called the related rates equation and you are given all you need to plug in numbers and solve for $s'$ at the moment when $x=50$.

 

[adjacent]

Then simply differentiate this equation with respect to $t$

Why do you differentiate the whole equation?

 

[CAF123]

Why do you differentiate the whole equation?

That is implicit differentiation, a technique employed when it is difficult (or impossible) to express a function f=f(t) so you can compute derivatives.

See http://tutorial.math.lamar.edu/Classes/CalcI/ImplicitDiff.aspx

How did you get on with using chain rule earlier?

 

[Ray Vickson]

Then I will need an initial distance,which is not given in the question.

Your expression $s_1 = \sqrt{0.12^2 + (6.5 t)^2}$ is OK for getting the answer: it just works backwards, starting at time $t=0$ from the base of the tower and walking away from the tower at speed 6.5 (km/hr); just be sure to change the sign at the end, since your distance will be increasing instead of decreasing (but the magnitude will be correct).

Alternatively, take $s_1 = \sqrt{0.12^2 + (h - 6.5 t)^2}$ with some unknown initial horizontal distance $h$. Since $h$ is constant, its derivative = 0 and so does not affect very much.

 

[adjacent]

I have solved the problem:
$s_1=\sqrt{120^2+s_2^2}$
I, now need to find the rate of change of $s_1$ with respect to time.
So I need the chain rule $$\frac{\text{d}s_1}{\text{d}t}=\frac{\text{d}s_1}{\text{d}s_2}.\frac{ds_2}{\text{d}t}$$
Now,
$$\frac{\text{d}s_1}{\text{d}s_2}=\frac{1}{2}(120^2+s_{2} ^2)^{\frac{-1}{2}}.2s_2$$
and $\frac{ds_2}{\text{d}t}=6.5$

After computing all these,I get 2.5km/h .Yay!

 

[LCKurtz]

Sorry I've been away from this thread for so long. If $x$ is the distance of the man from the base of the tower and $s$ is the distance from the top, you should have started with the original equation$$
x^2+120^2 =s^2$$and realize that $x$ and $s$ are functions of $t$. Then simply differentiate this equation with respect to $t$:$$
2xx' +0 = 2ss'$$This is called the related rates equation and you are given all you need to plug in numbers and solve for $s'$ at the moment when $x=50$.

Why do you differentiate the whole equation?

You have to differentiate both sides. Notice that by not solving for $s$ you never have to differentiate the square root. Just plug your numbers in $xx'=ss'$ giving $50(-6.5)=130s'$, so $s'=-2.5$. So $s$ is decreasing at $2.5$m/s. That's the same answer you have except for the sign.