Noora's questions at Yahoo Answers regarding linear approximations

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Discussion Overview

This discussion revolves around linear approximations in calculus, specifically addressing two problems: approximating the value of (26.98)^(4/3) and finding an approximate value of y for an implicitly defined function as x changes from 1 to 1.1. The scope includes mathematical reasoning and technical explanations related to calculus concepts.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents two calculus problems involving linear approximations, seeking assistance.
  • Another participant defines a function for the first problem and provides a step-by-step calculation to approximate (26.98)^(4/3) using linear approximation techniques.
  • The same participant implicitly differentiates a given relation to find dy/dx for the second problem and calculates an approximate value of y when x changes from 1 to 1.1.
  • Subsequent posts express gratitude for the assistance and suggest moving additional questions to a dedicated calculus sub-forum for better visibility and readability.

Areas of Agreement / Disagreement

There is no explicit disagreement noted in the discussion, but the initial poster's request for help indicates uncertainty about the problems. The responses provide calculations and methods without indicating any consensus on the correctness of the approaches.

Contextual Notes

The discussion does not clarify any assumptions made in the calculations or the implications of the approximations used. There may be limitations in the accuracy of the approximations due to the choice of values and methods applied.

Who May Find This Useful

Students or individuals interested in calculus, particularly those seeking help with linear approximations and implicit differentiation.

MarkFL
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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
 

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