# Related Rates (I think?)

1. Dec 13, 2009

### Linday12

1. The problem statement, all variables and given/known data
A magic stalk starts growing at a spot 10 metres from a 4 metre lamppost. The stalk grows 1 metre per hour. Find the rate at which the length 's' of the stalk's shadow is increasing at the instant when the height h of the stalk is 3 metres.

3. The attempt at a solution
First, I drew a diagram, 4m lamppost as left edge of a triangle, 3 metre height in the middle of the triangle, and 10+s base (10 meters from the beanstalk to the lamppost, and length 's' shadow. I know this part is right.

After this, I'm lost as to what I should do. I think it might be something with c^2=a^2+b^2, but the two equations seem to get very complex, so I must be doing something wrong, and I have no idea how I would combine them. (after which I assume I need the f'(x), because f'(x)=v'(t)=1, but really have no idea?

So any help in the right direction would be highly appreciated. Thanks.

2. Dec 13, 2009

### tiny-tim

Hi Linday12!

(try using the X2 tag just above the Reply box )
No, it's much simpler …

Hint: draw the horizontal line from the top of the magic stalk to the lamppost (so your big triangle is divided into two triangles and a rectangle).

3. Dec 13, 2009

### Linday12

Thanks, I'm studying for a final, and on the finals from prior years there seems to be questions like this one asked.

So, after drawing the other line, I can see two triangles in the diagram. I know the details of the one triangle, so if I remember correctly, I can find the details of the other one by using similar triangles. So, $$\dfrac{1}{10}$$ = $$\dfrac{3}{s}$$, and solving for that, I get s=30. Now, I must find the rate at which the length of s is increasing at the instant when the height h of the beanstalk is 3 metres. I assume this is the part where the calculus comes into play, and I don't simply say its growing at 10 metres an hour. So, I'm not exactly sure how to make the equation that I'm supposed to take the derivative of.

4. Dec 13, 2009

### tiny-tim

Yes, using similar triangles is correct, but keep the height of the stalk as a variable

or you won't have anything to differentiate with respect to!

So what is the length of the shadow when the height is h?

5. Dec 13, 2009

### Linday12

Ok, so taking that into account, I have $$\frac{1}{10}$$ = $$\frac{h}{s}$$, which then can be solved to become s=10h. s'=10, so the rate at which the length s is increasing is 10 metres per hour. I'm still not sure where the stalk growth of 1 metre per hour comes in at then.

6. Dec 13, 2009

### tiny-tim

erm … it's not 1 any more, is it?
Hint: you have three variables … h s and t

7. Dec 13, 2009

### Linday12

I think I understand what you mean (for a single variable course). And if I have 3 variables, that is one too many. So, since h is variable, then 1 must be variable t, and I can state it as 4-h. So, $$\frac{4-h}{10}$$ = $$\frac{h}{s}$$, so s=$$\frac{10h}{4-h}$$, and s'(h)=$$\frac{10}{4-h}$$ + $$\frac{10h}{(4-h)^2}$$. From here, subbing in the height of 3, I get 40. I think I am way off course.

8. Dec 13, 2009

### tiny-tim

Looks ok to me (but you should write ds/dh rather than s'(h), just to make it clear what you're differentiating with respect to, since ultimately you need ds/dt ).

What's wrong with 40 (except you haven't given the units)? Calculate the length when the height is 2 3 and 3.5.

(of course, in the next hour, the shadow will go an infinite distance! )

9. Dec 13, 2009

### Linday12

Interesting. So,
$$\frac{ds}{dh}$$ = $$\frac{10}{4-h}$$ + $$\frac{10h}{(4-h)^2}$$

where h is 3, the shadows rate is increasing at 40.... metres every 3 hours? That has to be wrong, since it doesn't really make any sense to me. I'm unsure of what the units would be.

And, when it approaches 4, the shadow does go to infinity. So, that part checks out.

And thank you for your help in finding the answer. It's very much appreciated, especially the day before a test.

10. Dec 13, 2009

### tiny-tim

Why do you have this aversion to using t (the time)?

(and where did that 3 come from?)

If you'd used t, you'd know what the units are!

11. Dec 13, 2009

### Linday12

Sorry, I'm not quite sure what I'm doing. I got the 3 for the height from the question, asking the shadows rate of increase at 3. It's a one variable calculus course, so I was avoiding t to keep it down to one variable, I'm not sure at all how to do differentiation with two variables. I could change the similar triangles to be: $$\frac{4-t}{10}$$ = $$\frac{h}{s}$$

Which then comes out to $$\frac{10h}{4-t}$$. I'm not sure how to do this part. I can sub the rate of 1 metre per hour in, and get s=10/3 for h=1 and t=1. Then I can do the same for h=3, and t=3, and get s=30, but if I have no idea how to differentiate the equation. I guess I could let h be t as well, since it will be the same as t, becomes:
$$\frac{ds}{dt}$$=$$\frac{10}{4-t}$$ + $$\frac{10t}{(4-t)^2}$$.

12. Dec 13, 2009

### tiny-tim

oooh, don't do that!

Put in the extra variable, and use the chain rule …

in this case, you want ds/dt, so that's ds/dh dh/dt (and you know dh/dt = 1 m/hr)

("one variable" means one at a time, but you can have chains of one-variable-at-a-time, in sequence )

13. Dec 13, 2009

### Linday12

Hmm, I am stumped. I'm not seeing how to do it at all.

14. Dec 13, 2009

### tiny-tim

But you've got it …

you had ds/dh, so you just multiply that by dh/dt (which is 1), and that gives you ds/dt in m/hr

$$\frac{ds}{dh}$$ = $$\frac{10}{4-h}$$ + $$\frac{10h}{(4-h)^2}$$, when the height, h, is 3, is equal to 40. So, then $$\frac{dh}{dt}$$ * 40, and solving the dh/dt for 1m/hr, makes the shadow have an increase of 40 m/hr at h=3. Hopefully I'm starting to get it.