Linear Approximation problems.....

**john11235** · Oct 26th 2008, 06:57 PM

1. Let $f(t)$ be the weight (in grams) of a solid sitting in a beaker of water. Suppose that the solid dissolves in such a way that the rate of change (in grams/minute) of the weight of the solid at any time can be determined from the weight using the formula $f'(t)=-3f(t)(5+(f(t))$ . If there are 4 grams of solid at $t=2$ , estimate the amount of solid one second later.

2. Use Linear Approximation to approximate $1.6^4$ as follows: Let $f(x)=x^4$ . The equation of the tangent line to $f(x)$ at $x=2$ can be written in the form $y=mx+b$ , where m=______ and b=______. Using this, the approximation of $1.6^4$ is _______.

3. The L.A. at x=0 of $Sqrt(3+5x)$ can be written as A+Bx where A=______ and B=______.

4. Use L.A. to approximate $1/1.003$ as follows: Let $f(x) = 1/x$ and find the equation of the tangent line to f(x) at a "nice" point near 1.003. Then use this to estimate $1/1.003$ .

5. The linearization at x=0 to $sin(6x)$ is A+Bx . Compute A and B.

For number 4 I've gotten that it's approximately 0.997017919, but it keeps saying that it's wrong. I have no clue how to figure out any of the others.

**TKHunny** · Oct 26th 2008, 07:07 PM

5) Have you considered actually answering the question? It asks for A and B and you have supplied only one value. How can that be right?

Think about the most basic algebra and geometry. Remember the Point Slope form of a line? (y-y0) = m(x-x0). That's all you need to know.

We have x0 = 0, that gives y0 = sin(6(0)) = sin(0) = 0

Further, if f(x) = sin(6x), we have f'(x) = 6cos(6x) and immediately m = f'(0) = 6cos(6(0)) = 6cos(0) = 6(1) = 6.

Putting it all together (y-y0) = m(x-x0) leads to (y-0) = 6(x-0). Simplifying a bit, we get y = 6x and we see that A = 0 and B = 6.

You show us the next one. 4 - maybe?

**john11235** · Oct 26th 2008, 07:50 PM

Originally Posted by TKHunny

5) Have you considered actually answering the question? It asks for A and B and you have supplied only one value. How can that be right?

Think about the most basic algebra and geometry. Remember the Point Slope form of a line? (y-y0) = m(x-x0). That's all you need to know.

We have x0 = 0, that gives y0 = sin(6(0)) = sin(0) = 0

Further, if f(x) = sin(6x), we have f'(x) = 6cos(6x) and immediately m = f'(0) = 6cos(6(0)) = 6cos(0) = 6(1) = 6.

Putting it all together (y-y0) = m(x-x0) leads to (y-0) = 6(x-0). Simplifying a bit, we get y = 6x and we see that A = 0 and B = 6.

You show us the next one. 4 - maybe?

Sorry, I hit the wrong button there.... I meant to say that I had tried number 4.

But now it's not making any sense.... I have $(Y-Yo)=(-1/1.003^2)(x-1.003)$ , but what's the Yo supposed to be?

**TKHunny** · Oct 27th 2008, 10:00 AM

You missed the really big hint! "Nice Point"

x0 is NOT 1.003. You should be using the "Nice Point" for x0 and y0.

Try x0 = 1 and 1/x0 = 1/1 = 1 = y0 so that you can approximate 1/1.003.

**john11235** · Oct 27th 2008, 12:28 PM

Originally Posted by TKHunny

You missed the really big hint! "Nice Point"

x0 is NOT 1.003. You should be using the "Nice Point" for x0 and y0.

Try x0 = 1 and 1/x0 = 1/1 = 1 = y0 so that you can approximate 1/1.003.

Oh ok.

I got it now. Thanks!

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