Please give me and example of what a calculus problem looks like?

Question

I really want to know this and have wanted to for a long time. But no one knows really. But lets see if you mathematicians can help me figure it out

Answer 1

Here's a simple one:

dy/dx = 3x^2 + 5x + 1

Answer 2

This is a problem from my Calculus book in chapter 7 section 3:
5. The solid lies between planes perpendicular to the axis at x= -1 and x=1. The cross sections perpendicular to the x-axis between these planes are circular disks whose diameter run from the parabola y= x^2 to the parabola y=2-x^2.

This kind of problem is for AP Calculus AB.

Answer 3

If Calculus I can be described in one word, it would be "derivative". An example of a word problem is below:

A parcel delivery service will deliver a package only if the length plus the girth (distance around) does not exceed 120 inches. Find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements.

Answer 4

You really wanna know, huh?

Ok. Let's say that there's a 25 foot ladder leaned against the wall. Some moron knocks it over and it starts sliding down the wall at such a rate that the bottom of the ladder is moving away from the wall at 18 inches per second. Find the rate at which the top of the ladder is sliding down the wall when the base of the ladder is 4.5 feet from the wall.

You'd set this up into sections of "want", "know", and "relate".

So
Want: dH/dt (essentially the rate at which the height is going down), when b (distance from base to wall) is 4.5 feet

Know: db/dt (the rate at which the base is moving away from the wall) = 18 inches, or 1.5 feet/sec

Relate. by the pythagorean theorem, the distance from the top of the ladder to the ground squared plus the distance from the bottom of the ladder to the wall squared would equal the length of the ladder sqaured.
so, H^2 + b^2 = 625

Then, take the derivative of both sides of the relate equation. The derivative is used to find instantaneous rates of change, as we are doing in this problem. It can be found by decreasing the exponent by one and making it the coeffecient. when taking the derivative of a constant, the answer is always 0.

so, the derivative in this case is:
2H*dh/dt + 2b*db/dt = 0
(the dh/dt)s and stuff are there because they are the derivative of the variable itself, which is necessary.

Before we substitute our information into the equation, we must find a value for the height of the top of the ladder.
we can use algebra for this.
the base to wall length is 4.5 feet, and the ladder is 25 feet long. so, using the Pythagorean theorem, we know that height squared + 4.5^2 = 625
eventually, you get the height as 24.59 feet.
so, now we can substitute.
2*24.59*dh/dt + 2*1.5*4.5 = 0
multiply this out and separate and get: 49.18dh/dt = -13.5
we are trying to solve for the rate at which the ladder is following, or dh/dt. given that that's the only variable left, we're in good shape.
then, divide by 49.18 to find the final answer:
dh/dt = -0.2745...feet/second

So the top of the ladder is moving to the ground at about 0.2745 feet per second, or 0.187 miles per hour.

see, once you know what you're doing, it's not that tough
:)

Answer 5

Lets say you are walking down a street at night at a certain speed. You pass a lamp where the bulb is, say, 20 feet high, and as you go along, you cast a shadow. You can use calculus to determine the length of the shadow as a function of time.

Answer 6

Its summer time and your local lake is below the necessary level for water skiing. The streams supply water at a rate of whatever gallons per day. But the damn provides power for your town and needs water to flow out of the lake to do so. Town has energy needs of whatever per day. The damn produces power at a rate of how much power per gallon of water flowing out of the lake. If you need to meet the energy needs of the town, in how many days can you go water skiing? You should be able to solve it, and you can look up realistic numbers to make it more real. And you get to go "water skiing". Good luck. Feel free to expand (maybe find out when the damn will break and how do you prevent that etc)

Answer 7

Use calculus to predict how much a rocket's acceleration rate might increase due to its becoming less massive as it uses fuel.

Answer 8

its monstrous.if ur being forced 2 take the class....well, u don wanna noe. no need 4 early heart attacks rite?