# Simple derivative problem

1. May 31, 2014

1. The problem statement, all variables and given/known data
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?
2. Relevant equations
Nah
3. 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

2. May 31, 2014

### 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.

3. May 31, 2014

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

4. May 31, 2014

### LCKurtz

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.

Last edited: May 31, 2014
5. May 31, 2014

### BiGyElLoWhAt

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

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

6. May 31, 2014

### BiGyElLoWhAt

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

7. Jun 1, 2014

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

Last edited: Jun 1, 2014
8. Jun 1, 2014

### CAF123

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

In this case, the distance of the man from the bottom of the tower is changing as t increases. So you have that sb = sb(t). But you know that the distance of the man from the top is a function of sb. So st = st(sb) = st(sb(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.

9. Jun 1, 2014

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.

10. Jun 1, 2014

### 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.

11. Jun 1, 2014

### CAF123

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.

12. Jun 1, 2014

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 *

13. Jun 1, 2014

### Fredrik

Staff Emeritus
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.

14. Jun 1, 2014

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

15. Jun 1, 2014

### Fredrik

Staff Emeritus
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).

Last edited: Jun 1, 2014
16. Jun 1, 2014

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

17. Jun 1, 2014

### CAF123

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.

18. Jun 1, 2014

### Fredrik

Staff Emeritus
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.

Last edited: Jun 1, 2014
19. Jun 1, 2014

### 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 $x=50$.

20. Jun 1, 2014