Calculus Problems Solved Step By Step

Calculus Problems Solved Step By Step-66
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.

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We could reason physically, or use the First Derivative Test, but we think it’s easiest in this case to use the Second Derivative Test.

Let’s quickly compute the second derivative, starting with the first derivative that we found above:\[ \begin A'(r) &= 4\pi r\, – 2V r^ \\[8px] A’^\prime(r) &= 4\pi \dfrac(r) -2V \dfrac\left(r^ \right) \\[8px] &= 4\pi -2V \left((-2)r^ \right) \\[8px] &= 4\pi \frac \end \]Since $r (as opposed to a maximum or saddlepoint), which is what we’re after: The minimum surface area occurs when $r = \sqrt[3]\,. Now that we’ve found the critical point that corresponds to the can’s minimum surface area (thereby minimizing the cost), let’s finish answering the question: The problem asked us to find the dimensions — the radius and height — of the least-expensive can.

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.

In this problem, for instance, we want to minimize the cost of constructing the can, which means we want to use .

Symbolic Essays Everyday Use - Calculus Problems Solved Step By Step

So let’s write an equation for that total surface area:\begin A_\text &= A_\text A_\text A_\text \[8px] &= \pi r^2 2\pi r h \pi r^2 \[8px] &= 2\pi r^2 2 \pi r h \end That’s it; you’re done with Step 2!first have to develop the function you’re going to maximize or minimize, as we did in Stage I above.Having done that, the remaining steps are exactly the same as they are for the max/min problems you recently learned how to solve.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.Constants come out in front of the derivative, unaffected: $$\dfrac\left[c f(x) \right] = c \dfracf(x) $$ For example, $\dfrac\left(4x^3\right) = 4 \dfrac\left(x^3 \right) =\, …$The derivative of a sum is the sum of the derivatives: $$\dfrac \left[f(x) g(x) \right] = \dfracf(x) \dfracg(x) $$For example, $\dfrac\left(x^2 \cos x \right) = \dfrac\left( x^2\right) \dfrac(\cos x) = \, …$\begin \dfrac(fg)&= \left(\dfracf \right)g f\left(\dfracg \right)\[8px] &= \Big[\text \times \text\Big] \Big[\text \times \text\Big] \end \begin \dfrac\left(\dfrac \right) &= \dfrac \[8px] &=\dfrac \end Many students remember the quotient rule by thinking of the numerator as “hi,” the demoninator as “lo,” the derivative as “d,” and then singing “lo d-hi minus hi d-lo over lo-lo” We’ll show more detailed steps here than normal, since this is the first time we’re using the Power Rule.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?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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