Question

Use differentials to find the approximate value:

Answer 1

\[203^{\frac{ 1 }{ 2 }} 18^{\frac{1}{4}}\]

Answer 2

Or, a different question:

Minimize:
\[x^2+y^2+z^2=7\]

Answer 3

Example: Approximate \(\sqrt{126}\) using differentials.
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\(\sqrt{126}\) is near \(\sqrt{121}=11\), so we will set
\(f(x)=\sqrt{x}\) with \(x=121\) and \(\Delta x=5\).

\(f(x+\Delta x)=f(x)+\Delta y~{\color{red}{=}}~f(x)+f'(x)\Delta x\)
\(f(x+\Delta x)=f(x)+(2\sqrt{x})^{-1}\Delta x\)
\(\sqrt{126}=\sqrt{121}+\frac{1}{2\sqrt{121}}(5)=11+\frac{5}{22}\)

Answer 4

The red equal sign is the sign that corresponds to where I made my approximation.
Precisely, we are approximating by assuming \(\Delta y \approx f'(x) \Delta x\).

(The concept is basically the same as with the tangent line approximation.)

Answer 5

So, in your case, you can approximate \(\sqrt{203}\), by setting \(\tiny \\[0.5em]\)
(1) \(f(x)=\displaystyle\sqrt{x}\), \(\tiny \\[0.5em]\)
(2) \(x=196\) (because \(\sqrt{196}=14\)), and \(\tiny \\[0.5em]\)
(3) \(\Delta x=7\) (because \(203-196=7\))

Answer 6

Then, you can approximate \(\sqrt[4]{18}\), by setting \(\tiny \\[0.5em]\)
(1) \(f(x)=\displaystyle\sqrt[4]{x}\), \(\tiny \\[0.5em]\)
(2) \(x=16\) (because \(\sqrt[4]{16}=2\)), and \(\tiny \\[0.5em]\)
(3) \(\Delta x=2\) (because \(18-16=2\))

Note that \(\Delta y=f'(x)\Delta x=\displaystyle \frac{\Delta x}{4\sqrt[4]{x^3}}= \frac{2}{4\sqrt[4]{16^3}}= \frac{1}{2\sqrt[4]{(2^4)^3}}= \frac{1}{16}\).

Answer 7

Then, as you approximate \(\sqrt{203}\) and \(\sqrt[4]{18}\) individually multiply these individual approximations to get the approximation for the product.

Answer 8

Next, I will minimize \(z^2+x^2+y^2=13\).

The \(z\)-value is the smallest when \(z=-\sqrt{13}\),
because \(\sqrt{13}\) is the smallest value on the sphere.
So we already know what the minimum is.

So we need \(x^2+y^2=0\).
The only way for \(x\) and \(y\) to meet this equation is \((x,y)=(0,0)\).
So the absolute minimum is \(x^2+y^2+z^2=13\).

(No calculus here.)

Answer 9

The absolute minimum occurs \((0,0,-\sqrt{13})\) ....
For any sphere with radius \(\delta\), the minimum is \((0,0,-\delta)\),
and the maximum is \((0,0,\delta)\).

Answer 10

well, for any such sphere that is centered at the origin of course.

Answer 11

Im here btw, just working this out alongside you

Answer 12

I am pretty sure you understand the sphere thing.
You know what the sphere looks like precisely, so there you know the maximum and the minimum of it ....

Answer 13

Do you have any questions about the approximation using differentials?

Answer 14

\(\color{blue}{\text{Originally Posted by}}\) @SolomonZelman
Example: Approximate \(\sqrt{126}\) using differentials.
-------------------------------------------------

\(\sqrt{126}\) is near \(\sqrt{121}=11\), so we will set
\(f(x)=\sqrt{x}\) with \(x=121\) and \(\Delta x=5\).

\(f(x+\Delta x)=f(x)+\Delta y~{\color{red}{=}}~f(x)+f'(x)\Delta x\)
\(f(x+\Delta x)=f(x)+(2\sqrt{x})^{-1}\Delta x\)
\(\sqrt{126}=\sqrt{121}+\frac{1}{2\sqrt{121}}(5)=11+\frac{5}{22}\)
\(\color{blue}{\text{End of Quote}}\)

Is the delta y here supposed to be delta x?

Answer 15

Oh, no it shouldn't.

Answer 16

that was the only thing that threw me off.