VCE Methods Question Thread!

[StressedAlready]

How will it give the required graph? We want height against time... if we derive the volume formula, won't we get dV/dx?

[garytheasian]

How will it give the required graph? We want height against time... if we derive the volume formula, won't we get dV/dx?

Oh woops my bad didnt read the question properly. So i guess it requires you to dx/dt=dv/dt X dx/dv which you have found out if my previous step is followed all you need to do is flip it and then multiply it by dv/dt which is given in the question.

[StressedAlready]

For this, we would use dv/dt=-1000cm^3/min, right?

[garytheasian]

For this, we would use dv/dt=-1000cm^3/min, right?

yup

[Floatzel98]

yup

Is the graph of dh/dt the same as the graph of h against t, h(t)? I think the question was the graph of the water height against time. Would we have to integrate dh/dt still to find the equation? If we are supposed to find that, how can we solve for the constant?

[garytheasian]

Is the graph of dh/dt the same as the graph of h against t, h(t)? I think the question was the graph of the water height against time. Would we have to integrate dh/dt still to find the equation? If we are supposed to find that, how can we solve for the constant?

Facepalm didnt read the question fully again, hmmm at this point im not sure but i suspect that we assume it is a definite integral otherwise i have no clue.

[kinslayer]

You need an initial value for x/h to do this question. Do we assume that the cone is full of water before you start? Just read StressedAlready's post.

So you are given an initial value at t=0 the water is at a height of 40cm. Go nuts!

[garytheasian]

If you don't do it graphically how would you know x>-2? What is there to indicate that the less than becomes a greater than?

[kinslayer]

If you don't do it graphically how would you know x>-2? What is there to indicate that the less than becomes a greater than?

You get your first clue by solving for zero: you know that f'(x) touches the x-axis at x =-2 and x = 0. Moreover it is a quadratic, so it must cross the x-axis at those two points.

From here, the quickest way to the solution is to note that f'(-1) is negative (alternatively, f(-3) is positive, or f(1) is positive). You only need to check one point out of any of the three intervals x < -2, -2 < x < 0, or x > 0. Either way, it follows that it is negative between -2 and 0 and positive elsewhere.

If you want to use the inequality, you need to use the fact that a positive number times a positive number is positive, as is a negative number times a negative number. If you want the product of two numbers to be negative, then one of them must be negative and the other must be positive.

We can definitely say that if x > 0, then x and x+2 are both positive. If x < -2, then x and x+2 are both negative. If x is less than zero, but greater than -2, x is negative, but x+2 is positive.

So what we want is .

[garytheasian]

You get your first clue by solving for zero: you know that f'(x) touches the x-axis at x =-2 and x = 0. Moreover it is a quadratic, so it must cross the x-axis at those two points.

From here, the quickest way to the solution is to note that f'(-1) is negative (alternatively, f(-3) is positive, or f(1) is positive). You only need to check one point out of any of the three intervals x < -2, -2 < x < 0, or x > 0. Either way, it follows that it is negative between -2 and 0 and positive elsewhere.

If you want to use the inequality, you need to use the fact that a positive number times a positive number is positive, as is a negative number times a negative number. If you want the product of two numbers to be negative, then one of them must be negative and the other must be positive.

We can definitely say that if x > 0, then x and x+2 are both positive. If x < -2, then x and x+2 are both negative. If x is less than zero, but greater than -2, x is negative, but x+2 is positive.

So what we want is .

i dont quite understand this "From here, the quickest way to the solution is to note that f'(-1) is negative (alternatively, f(-3) is positive, or f(1) is positive). You only need to check one point out of any of the three intervals x < -2, -2 < x < 0, or x > 0. Either way, it follows that it is negative between -2 and 0 and positive elsewhere." but i do get the second part so thank you!

[kinslayer]

i dont quite understand this "From here, the quickest way to the solution is to note that f'(-1) is negative (alternatively, f(-3) is positive, or f(1) is positive). You only need to check one point out of any of the three intervals x < -2, -2 < x < 0, or x > 0. Either way, it follows that it is negative between -2 and 0 and positive elsewhere." but i do get the second part so thank you!

The idea is that since it is a quadratic and it has two zeroes, it must change sign at each point. There's simply no other way for a second degree polynomial to have two zeroes.

So, we have two points where it changes sign. To the left of the leftmost zero, it is one sign. Then after the leftmost zero but before the rightmost zero, it is the other sign. Then after the rightmost zero, it is the same sign as it was to the left of the leftmost zero.

That was kind of a mouthful, but basically a quadratic has one sign between the two zeroes and the opposite sign on either side. This is a consequence of the fact that it changes sign at both zeroes.

If the function were a cubic or some other other higher degree polynomial then that is certainly not true. It's a quadratic though and this is something you should know about quadratics

[knightrider]

When finding the average value of a function.

does the final answer have any units.
if not, why?

Also in the final answer do you have to say therefore y av=blah ?

[garytheasian]

When finding the average value of a function.

does the final answer have any units.
if not, why?

Also in the final answer do you have to say therefore y av=blah ?

im sure you need to and yes

[nerdgasm]

I think that the average value of a function (by default) is unitless, because values of a function are unitless, and taking the average of these values shouldn't require putting some units in front of it. For example, if I asked you what the average of 1, 2, 3, 4 and 5 is, the answer would be 3, not 3 units.
Another way to look at it is that the average value of a function in the interval [a, b] is the value of the constant function which when integrated over [a,b] gives you the same result as integrating the original function over [a, b]. This is to show that the average value is just a number - you wouldn't write f(x) = 3 cm^2; you would write f(x) = 3.

In some sense, it's the 'real-world' information provided along with the function that determines the units. Consider a function, f(x) = x^3, dom(x) = (0, inf). While I can use this function to calculate the volume of a cube of side length x, that doesn't really mean that f(2) = 8 units^3. This is because the function f(x) = x^3, dom(x) = (0, inf) doesn't really have an intrinsic real-world meaning until we place some on it (by noticing that this function would be good at letting us calculate volumes of cubes, for example). Even though many functions have been inspired by real-world phenomena, at the end of the day, the functions are not intrinsically linked to them - they can be used as context-free mathematical objects, and hence do not require units. As another example, the average velocity of a car that uniformly accelerates at 1m/s^2 from 0m/s to 10m/s will be 5m/s over the first 10 seconds. But the average value of y = x over [0, 10] is just 5.

I suppose as a final comment, consider a (single-variable) function that maps from the real numbers to the real numbers (f: R -> R, f(x) = blah). What this means is that the function takes in a real number within its domain (this is the point of the first R), and spits out another real number (this is the point of the second R). What I'm trying to get at is that the function just spits out numbers without any units. Hence, the values of functions are unitless, and hence the average value of a function is unitless too.

[I_I]

Hi, guys! I have a SAC soon on Calculus and I'm having trouble understanding a specific type of questions on optimisation problems.
I;m pretty okay at most optimisation problems but questions that go "Find the area of the largest "so and so" that can fit inside "so and so" or "Find the dimensions of a largest triangle that can be inscribed in "so an so" " and such make me go crazy

I pulled out a couple of questions that I don't know and was wondering if anybody would be so kind to go through them, step by step

1. Find the area of the largest rectangle that fit inside the a right angled triangle with height 3cm and base 4 cm (Ans: 3cm squared)

2. Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r. (answer: base: sqr(3), height 3r/2)

3. A right circular cylinder is inscribed in a cone with height h and base radius . Find the largest possible volume of such a
cylinder.

I would really appreciate it if someone could answer!