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# Tutor profile: Chris F.

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Chris F.
Purveyor of Mathematics
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## Questions

### Subject:Pre-Calculus

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Question:

Suppose a block of concrete has a square base that measures $$x$$ inches on each side, the height of the block is $$y$$ inches, and the total volume of the block is 4,000 cubic inches. Write a function that determines the surface area of the block in terms of the length of the sides $$x$$.

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Chris F.

First, recall that the volume $$V$$ of the block can be written as, \begin{align*} V&=x^2y.\\ \end{align*} Next, recall that the surface area $$A_{s}$$ of the block can be written as \begin{align*}A_{s}=2x^2+4xy.\end{align*} Now we will solve the volume equation for $$y$$ as follows, \begin{align*} V&=x^2y\\ \frac{1}{x^2}\cdot V&=\frac{1}{x^2}\cdot x^2y\\ \frac{1}{x^2}\cdot V&=y.\\ \end{align*}Using the equation $$y=\frac{1}{x^2}\cdot V,$$ we will substitute $$y$$ into our surface area formula and simplify as follows, \begin{align*} A_{s}&=2x^2+4xy\\ &=2x^2+4x\bigg(\frac{1}{x^2}\cdot V\bigg)\\ &=2x^2+4\bigg(\frac{1}{x}\cdot V\bigg)\\ A_{s}&=2x^2+\frac{4V}{x}. \end{align*} Now we substitute the given volume of the block into the equation $$A_{s}$$ and get \begin{align*} A_{s}&=2x^2+\frac{4(4000)}{x}\\ &=2x^2+\frac{16000}{x}.\\ \end{align*} With our formula for the surface area now written only in terms of $$x,$$ we can write a function that determines the surface area of our given block in terms of the lengths of its sides as \begin{align*}A(x)=2x^2+\frac{16000}{x}.\end{align*}

### Subject:Calculus

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Question:

Let $$\vec{f},\vec{g}: \mathbb{ R} \to\mathbb{ R}^n$$ be differentiable, and prove $$\frac{d}{dt} \left\langle\vec{f}\vec{g}\right\rangle$$=$$\left\langle \frac{d \vec{f}}{dt}, \vec{g} \right\rangle$$+$$\left\langle \vec{f}, \frac{d \vec{g}}{dt} \right\rangle$$.

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Chris F.

Let $$\vec{f},\vec{g}: \mathbb{R} \to \mathbb{R}^n$$ be differentiable. Then \begin{align*} \frac{d}{dt}\left\langle\vec{f}\vec{g}\right\rangle&=\frac{d}{dt}(f_1g_1+f_2g_2+...+f_ng_n)\\ &=\frac{d}{dt}\bigg(\sum_{i=1}^{n}f_ig_i\bigg)\\ &=\sum_{i=1}^{n}\frac{d}{dt}(f_ig_i)\qquad\qquad\qquad\qquad\text{(term by term differentiation)}\\ &=\sum_{i=1}^{n}\bigg(\frac{df_i}{dt} g_i+f_i \frac{dg_i}{dt}\bigg)\;\qquad\qquad\text{(product rule)}\\ &=\sum_{i=1}^{n}\bigg(\frac{df_i}{dt} g_i\bigg)+\sum_{i=1}^{n}\bigg(f_i \frac{dg_i}{dt}\bigg)\;\quad\text{(commutative property)}\\ &=\left\langle\frac{d\vec{f}}{dt}{ \vec{g}}\right\rangle+\left\langle \vec{f}\frac{d\vec{g}}{dt}\right\rangle. \end{align*}

### Subject:Algebra

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Question:

Find the solutions to the quadratic equation $$x^2+x-1=0.$$

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Chris F.

First, recall that the quadratic formula states that if $$ax^2+bx+c=0,$$ then \begin{align*}x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.\end{align*} In our problem we see that $$a=1, b=1, \text{ and } c=-1.$$ Therefore, we substitute our known values for $$a, b, c$$ into the quadratic formula and simplify as follows, \begin{align*}x=\frac{-(1)\pm\sqrt{(1)^2-4(1)(-1)}}{2\cdot (1)}=\frac{-1\pm\sqrt{1+4}}{2}=\frac{-1\pm\sqrt{5}}{2}.\end{align*}Therefore, \begin{align*} x&=\frac{\sqrt{5}-1}{2}\quad\text{ or }\quad x=\frac{-(1+\sqrt{5})}{2}. \end{align*}

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