# Calculus Problems Solved Step By Step

For instance, a few weeks ago you could have gotten this as a standard max/min homework problem: You would probably automatically find the derivative $A'(r)$ (which you could equivalently write as $\dfrac)$, then find the critical points, then determine whether each represents a maximum or a minimum for the function, and so forth.That’s exactly what we’re now going to do in Stage II. ) Remember that is just a constant — it’s some number, like 355.

Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself.

Jump down this page to: [Power rule, $x^n$] [Exponential, $e^x$] [Trig derivatives] [Product rule] [Quotient rule] [Chain rule]$$\frac\text = 0 \quad \frac \left(x\right) = 1$$ $$\frac \left(x^n\right) = nx^$$\begin \frac\left( e^x \right) &= e^x &&& \frac\left( a^x \right) &= a^x \ln a \\ \\ \end\begin \frac\left(\sin x\right) &= \cos x &&& \frac\left(\csc x\right) &= -\csc x \cot x \\ \\ \dfrac\left(\cos x\right) &= -\sin x &&& \frac\left(\sec x\right) &= \sec x \tan x \\ \\ \dfrac\left(\tan x\right) &= \sec^2 x &&& \frac\left(\cot x\right) &= -\csc^2 x \end Notice that a negative sign appears in the derivatives of the co-functions: cosine, cosecant, and cotangent.

We’ve labeled the can’s height Having drawn the picture, the next step is to write an equation for the quantity we want to optimize.

Most frequently you’ll use your everyday knowledge of geometry for this step.

The first stage doesn’t involve Calculus at all, while by contrast the second stage is just a max/min problem that you recently learned how to solve: single variable. You’ll use your usual Calculus tools to find the critical points, determine whether each is a maximum or minimum, and so forth. In Optimization problems, always begin by sketching the situation. If nothing else, this step means you’re not staring at a blank piece of paper; instead you’ve started to craft your solution.

The problem asks us to minimize the cost of the metal used to construct the can, so we’ve shown each piece of metal separately: the can’s circular top, cylindrical side, and circular bottom.Typical phrases that indicate an Optimization problem include: Before you can look for that max/min value, you first have to develop the function that you’re going to optimize.There are thus two distinct Stages to completely solve these problems—something most students don’t initially realize [Ref]. Now maximize or minimize the function you just developed.We’ve already found the relevant radius, $r = \sqrt[3]\,.$To find the corresponding height, recall that in the Subproblem above we found that since the can must hold a volume of liquid, its height is related to its radius according to $$h = \dfrac\,.$$ Hence when $r = \sqrt[3]\,,$ $\begin h &= \frac\,\frac \[8px] &= \frac\,\frac \[8px] &= \frac\,\frac \[8px] &= 2^\frac \[8px] h &= 2^\sqrt[3] \quad \triangleleft \end$ The preceding expression for is correct, but we can gain a nice insight by noticing that $^ = 2 \cdot\frac$$and so $\begin h &= 2^\sqrt[3] \[8px] &= 2 \cdot\frac\,\sqrt[3] \[8px] &= 2 \sqrt[3] = 2r \end$ since recall that the ideal radius is$r = \sqrt[3]\,.$Hence the ideal height (height and radius) will minimize the cost of metal to construct the can?Above, for instance, our equation for$A_\text$has two variables, We can now make this substitution$h = \dfrac$into the equation we developed earlier for the can’s total area: $\begin A_\text &= 2\pi r^2 2 \pi r h \[8px] &= 2\pi r^2 2 \pi r \left( \frac\right) \[8px] &= 2\pi r^2 2 \cancel \cancel \left(\frac\right) \[8px] &= 2\pi r^2 \frac \end$We’re done with Step 3: we now have the function in terms of a single variable, , and we’ve dropped the subscript “total” from$A_\text$since we no longer need it.This also concludes Stage I of our work: in these threes steps, we’ve developed the function we’re now going to minimize!” We have provided those two dimensions, and so we are done.$\checkmark$We’ve now illustrated the steps we use to solve every single Optimization problem we encounter, and they always work. The first does not involve Calculus at all; the second is identical to what you did for max/min problems.Recall that$\dfrac\left(x^n\right) = nx^.\$ $\begin \dfrac\left(\sqrt[3]\, – \dfrac \right) &= \dfrac \left(x^ \right) – \dfrac \left(x^ \right) \[8px] &= \left(\fracx^ \right) – \left(-\frac x^ \right) \[8px] &= \frac x^ \fracx^ \[8px] &= \frac\frac \frac \frac \quad \cmark \end$ We’ll learn the “Product Rule” below, which will give us another way to solve this problem.For now, to use only the Power Rule we must multiply out the terms.

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