MHB Noora's questions at Yahoo Answers regarding linear approximations

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Noora seeks assistance with linear approximations in calculus, specifically for approximating (26.98)^(4/3) and finding the value of y in an implicit function as x changes from 1 to 1.1. The first approximation uses the function f(x) = x^(4/3) and calculates the derivative, leading to an estimated value of approximately 80.92. For the implicit function, after differentiating and applying the linear approximation, the estimated value of y at (1.1) is found to be 0.4. The discussion encourages posting additional calculus questions in a dedicated forum for better visibility and assistance.
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Here are the questions:

Help with CALCULUS approximations PLEASE?!?


a) Using an appropriate linear approximation approximate (26.98)^(4/3)

b) Suppose a function is defines implicitly by ((x^2)(y^2)) - 3y = 2x^4 - 4. Find the approximate value of y where (x, y) starts as (1,1) and x changes from 1 to 1.1

THANK YOU!

I have posted a link there to this topic so the OP can see my work.
 
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Hello Noora,

a) Let's define:

$$f(x)=x^{\frac{4}{3}}$$

Hence:

$$f'(x)=\frac{4}{3}x^{\frac{1}{3}}$$

For a small $\Delta x$, we know:

$$\frac{\Delta f}{\Delta x}\approx\frac{df}{dx}$$

Multiplying through by $\Delta x$ and using $Delta f=f\left(x+\Delta x \right)-f(x)$ we have:

$$f\left(x+\Delta x \right)-f(x)=\frac{df}{dx}\Delta x$$

$$f\left(x+\Delta x \right)=\frac{df}{dx}\Delta x+f(x)$$

Now, choosing:

$$x=27,\,\Delta x=-0.02$$

and using our function definition, we obtain:

$$(26.98)^{\frac{4}{3}}\approx\frac{4}{3}(27)^{\frac{1}{3}}(-0.02)+(27)^{\frac{4}{3}}=81-0.08=80.92$$

b) We are given the implicit relation:

$$x^2y^2-3y=2x^4-4$$

Implicitly differentiating with respect to $x$, we find:

$$x^2\cdot2y\frac{dy}{dx}+2xy^2-3\frac{dy}{dx}=8x^3$$

$$\frac{dy}{dx}\left(2x^2y-3 \right)=8x^3-2xy^2$$

$$\frac{dy}{dx}=\frac{2x\left(4x^2-y^2 \right)}{2x^2y-3}$$

Now, for a small $\Delta x$, we have:

$$\frac{\Delta y}{\Delta x}\approx\frac{dy}{dx}$$

$$y\left(x+\Delta x \right)\approx\frac{dy}{dx}\cdot\Delta x+y(x)$$

Using the given:

$$(x,y)=(1,1),\,\Delta x=0.1$$

we find:

$$y(1.1)\approx\frac{2(1)\left(4(1)^2-(1)^2 \right)}{2(1)^2(1)-3}\cdot0.1+1=0.4$$
 
Thank you SO much for your time and help! Can you please help me out with my other calculus-related questions on yahoo answers that i posted recently? Thank you!
 
ayahouyee said:
Thank you SO much for your time and help! Can you please help me out with my other calculus-related questions on yahoo answers that i posted recently? Thank you!

You're welcome! I'm glad you joined us here! :D

I would recommend you post them here in our Calculus sub-forum. That would be much easier than for me to try to do a search for your topics at Yahoo. :D

Also being able to use $\LaTeX$ here makes for much more readable help.
 
Thanks! I just made a new thread with my questions :D
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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