# Calculus Question

1. May 29, 2005

### Keyboard

Ok, so I got this assignment for Calculus class, and I'm having a hell of a time figuring it out, so is my tutor. Here are the 2 questions:

1.) Grain is emptying from the bottom of a funnel shaped hopper at 1.2m^3/min. If the diameter of the top of the hopper is 5m and the sides make an angle of 30(degrees) with the vertical. Determine the rate at which the level of grain in the hopper is changing, when it is half full.

2.) A cyclist travelling 20km/h rides east for one half hour, then north for one hour. How fast is the distance from the cyclist to the starting point increasing when the cyclist has been riding for 1.5h at 20km/h?

Any help would be greatly appreciated! I really need to get these questions right, I REALLY APPRECIATE THIS!

P.S. I would have asked my teacher for help, but since it's an assignment, were not allowed to ask for help from her, and my tutor didn't help much, I would really really appreciate if someobody could help me.

Thank you.

2. May 29, 2005

### Kruger

Say what are your ideas to solve that?

3. May 29, 2005

### jma2001

Does "half full" mean half the height, or half the volume?

4. May 29, 2005

### Keyboard

It doesn't specify.

5. May 29, 2005

### Staff: Mentor

When are you allowed to ask for help from her?

6. May 29, 2005

### bross7

Ah, rate problems.

When doing these problems always start with a diagram (you don't need to post an image of it). What you have in the first question as a funnel is an upsidedown cone. Draw this and put down your givens.

What is the equation for the volume of a cone?

7. May 29, 2005

### HallsofIvy

Staff Emeritus
Only when you don't need it!

1) The funnel is a cone, so start by drawing a pricture! You are told that the height of the funnel is 5 meters and that the sides make an angle of 30 degrees with the vertical. From your picture it should be easy to see that the radius (of the top circle) is height times tan(30). Now write down the equation for volume of a cone as a function of height and radius (I'll be that's in your text). Replacing r by h*tan(30) gives you V as a function of h only. Since you want "rate at which the level of grain in the hopper is changing", i.e. dh/dt, and you are told "Grain is emptying from the bottom of a funnel shaped hopper at 1.2m^3/min", i.e. dV/dt= -1.2, differentiate both sides of the equation with respect to time t and solve for dh/dt. Of course, that will depend upon both h and V. I would interpret "half full" as meaning half the volume (although I might also make that explicit in my answer) so you will need to calculate the initial volume of the funnel and find half of that. You will also need to calculate what h is when V is half its initial value.

2) How far east did the cyclist go in 1/2 hour? How far north did the cyclist go in the hour? Now draw a picture showing his initial position, his route east and his route n and connect his initial point to his final point. Do you see a right triangle? Do you remember the Pythagorean theorem? The east leg of the triangle has constant length, of course (the distance you calculated for my first question). As long as he is riding north, the length of the north leg is 20t where t is the time since he turned north. Can you write the length of the hypotenuse in terms of t? Differentiate both sides of that with respect to t to find the rate of change.

(The real question is "why would a motorcylist ride 1 and 1/2 hours at 20 stinking km per hour!?")

8. May 29, 2005

### jma2001

But that is not quite correct, the problem states "the diameter of the top of the hopper is 5m and the sides make an angle of 30(degrees) with the vertical." So you are given the radius of the top circle (1/2 diameter = 2.5m), and you need to use tan(30) to find the height.

9. May 29, 2005

### HallsofIvy

Staff Emeritus
Ouch! Thanks for catching that.

10. Jun 1, 2005

### Keyboard

Ok, so my friend helped me get to this step:

This is what one of my friends told me to do:
How do I solve for C'?

Thanks!

11. Jun 1, 2005

### HallsofIvy

Staff Emeritus
First, that is not "explicit" differentiation- it is "implicit" differentiation. "Explicit" differentiation would, I guess, be just regular differentiation! If you don't see why, look up "implicit" and "explicit" in a dictionary.

As far as solving 2C C'= 2AA'+ 2BB', if you are taking calculus, you should be embarassed by this: divide both sides of the equation by 2C!

C'= (A/C)A'+ (B/C)B'

BUT! Assuming this is the second problem you originally gave, at the moment in question, the bicyclist has alread gone east at 20km/h for 1/2 hour- that is 10 km-
and turned north. In other words the Pythagorean theorem in this case is just
C2= 10o+ B2 where B is the distance north. Differentiating that with respect to t, 2C C'= 2B B' so C'= (B/C)B'. (That's the same as taking A= 10, A'= 0 in the previous equation.)
At the moment in question the cyclist has riding north at 20 km/h for one and a half hours: B= 30 km and B'= 20 km/h. Of course, C is given by C2= 100+ 900= 1000.