# Noora's questions at Yahoo! Answers regarding linear approximations

#### MarkFL

Staff member
Here are the questions:

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.

#### MarkFL

Staff member
Hello Noora,

a) Let's define:

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

Hence:

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

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

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

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

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

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

Now, choosing:

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

and using our function definition, we obtain:

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

b) We are given the implicit relation:

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

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

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

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

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

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

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

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

Using the given:

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

we find:

$$\displaystyle y(1.1)\approx\frac{2(1)\left(4(1)^2-(1)^2 \right)}{2(1)^2(1)-3}\cdot0.1+1=0.4$$

#### ayahouyee

##### New member
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!

#### MarkFL

Also being able to use $\LaTeX$ here makes for much more readable help.