# Calculus Questions: Help Solve 1.2m^3/min & 20km/h Challenges!

• Keyboard
In summary: You have been given that A= 1/2 (hour east) and B= 1 hour (north). You have been told that A'= 20 km/hr and B'= 20 km/hr. C is not time- it is the distance from the starting point to the bicycle. I don't know where you got that number 1.11. If we take C to be the distance from the starting point to the bicycle, then C'= dC/dt and C^2= A^2+ B^2 so, by implicit differentiation, 2C C'= 2A A'+ 2B B'. But you don't know enough to solve for C' until you know the
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 traveling 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.

Say what are your ideas to solve that?

Keyboard said:
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.
Does "half full" mean half the height, or half the volume?

It doesn't specify.

Keyboard said:
I would have asked my teacher for help, but since it's an assignment, were not allowed to ask for help from her

When are you allowed to ask for help from her?

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?

jtbell said:
When are you allowed to ask for help from her?
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!?")

HallsofIvy said:
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).
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.

Ouch! Thanks for catching that.

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

This is what one of my friends told me to do:
this is EXPLICIT defferentiation! derevie expilictly with respect to C, so you get:

2*C*C' = 2A * A' + 2B*B' (second part my be wrong,so check it, but you can't expect to get the whole solution, right ?)

A' and B' are the speed he traveled at (20km/hr), A is time, B is time, C is time (1.11) and C' is what you are looking for.

How do I solve for C'?

Thanks!

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.

## 1. What is the formula for solving 1.2m^3/min and 20km/h challenges?

The formula for solving these types of challenges is to first identify the given values and units and then use the appropriate conversion factors to convert the units to a common measurement. In this case, we would convert 1.2m^3/min to m^3/hour and 20km/h to m/hour. Once the units are in the same measurement, we can then use basic algebraic equations to solve the problem.

## 2. How do I approach solving these types of calculus questions?

The best way to approach solving any calculus question is to first understand the given values and units, and then use the appropriate formulas and conversion factors to solve for the unknown variable. It is also helpful to draw a diagram or graph to visualize the problem and better understand the relationship between the given values.

## 3. What are some common mistakes to avoid when solving these types of challenges?

One common mistake to avoid is not properly converting the units to a common measurement. This can lead to incorrect solutions and answers. Additionally, it is important to double check your calculations and make sure you are using the correct formulas for the given problem.

## 4. How can I check my answer to make sure it is correct?

You can check your answer by plugging it back into the original equation and making sure it satisfies the given values and units. You can also use online calculators or ask a classmate or teacher to review your solution and provide feedback.

## 5. Can I use calculus to solve real-world problems?

Yes, calculus is a powerful mathematical tool that can be used to solve a wide range of real-world problems, including those involving rates, distances, and other physical quantities. It is used in fields such as physics, engineering, economics, and many more.

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