1. ## Linear Approximation problems.....

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.

2. 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?

3. 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?

4. 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.

5. 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!