Recent content by Elias Waranoi

In that case, k = xyz, w = x + y + z = k/(yz) + k/(xz) + k/(xy) ∂w/∂x = -k/(x2z) - k/(x2y) = -y/x - z/x = 0 ∴ y = -z ∂w/∂y = -k/(y2z) - k/(y2x) = -x/y - z/y = 0 ∴ z = -x ∂w/∂z = -k/(z2y) - k/(z2x) = -x/z - y/z = 0 ∴ x = -y Now this confuses me, because y = -z therefore x = -y = z. Now I have x...

This chapter was the first and only chapter on partial differentation. They showed that to find the maxima and minima of a multivariable function, the derivative of all partial derivatives have to be zero.

Never heard of it. I'm working with the Calculus made easy book and I haven't encountered it there.

Homework Statement Verify that the sum of three quantities x, y, z, whose product is a constant k, is maximum when these three quantities are equal. Homework Equations w = x + y + z k = x * y * z The Attempt at a Solution Assuming that my understanding of the question is correct i.e. that we...

So like this: ln y = ln(bεax) = ln(b) + ax ln(ε) and because ln(ε) = 1 then ln y = ln(b) + ax. Thank you very much

I know that deriving y = bεax gives dy/dx = abεax, I also know that the inverse of dx/dy equals dy/dx, so how come I don't get dy/dx = abεax when I do the following: ln(y) = ln(bεax) = ax ln(bε) -> x = ln(y)/(a ln(bε)) -> dx/dy = 1/(ay ln(bε)) -> dy/dx = ay ln(bε) = abεax ln(bε) ≠ abεax What...

But isn't v = ln(ax) = x ln(a) dv/dx = ln(a) I can get that dv/dx is ax ln(a) if v = ax so: v = ax -> ln(v) = ln(ax) = x ln(a) (1/v)(dv/dx) = ln(a) -> dv/dx = v ln(a) = ax ln(a) But v = ln(ax) so I can't get anything but ln(a)

I went through an example question that showed me how to solve the question but I'm not sure if I've misunderstood something or if they did a mistake. Question: Derivate y = (1/ax)ax ln(y) = ln( (1/ax)ax ) = ax( ln(1) - ln(ax) ) = -ax ln(ax) (1/y)(dy/dx) = -ax * ax ln(a) - a * ln(ax) dy/dx =...

From the book Calculus made easy: "This process of growing proportionately, at every instant, to the magnitude at that instant, some people call a logarithmic rate of growing." From Wikipedia: "Exponential growth is feasible when the growth rate of the value of a mathematical function is...

Sorry, bad title. What I'm looking for is the maximum in the curve of the growth of area for a cylinder inscribed in a cone. Not the maximum area. What I'm hung up on the result of my derivation dA/drc = 0 = 2πr. If for example my cone had the radius r of 1 meter than my derivation would give...

Homework Statement Inscribe in a given cone, the height h of which is equal to the radius r of the base, a cylinder (c) whose total area is a maximum. Radius of cylinder is rc and height of cylinder is hc. Homework Equations A = 2πrchc + 2πrc2 The Attempt at a Solution r = h ∴ hc = r - rc A =...

QC = paVa * ln(Vb/Va) + VbCv(pc - paVa/Vb)/R QH is heat the leaves the working substance so it should be negative. Apparently K = |QC|/( |QH| - |QC| ), thanks for the help I got the correct answer now.

Homework Statement The pV-diagram in Fig. P20.51 (See attached file) shows the cycle for a refrigerator operating on 0.850 mol of H2. Assume that the gas can be treated as ideal. Process ab is isothermal. Find the coefficient of performance of this refrigerator. Homework Equations K = QC/|QH -...

h/(h+y) - 1 = h/(h+y) - (h+y)/(h+y) = (h - (h+y))/(h+y) = (h - h - y)/(h+y) = -y/(h+y) oops, you're right. But what am I supposed to do now?

I don't even know what y<<h is. And simplifying h/(h+y)-1 gives me (h-y)/(h+y)