1. ## What is being asked?

I have a 4 part problem that I am unsure what is being asked. I have shown what i have worked out.

$Lety = f(x) = \sqrt{x}$

a) Find the differential of y

$dy=\frac{1}{2\sqrt{x}} dx$

b) Calculate dy when x=4 and delta x = 0.01. Interpret.

$dy= \frac {1}{4} * \frac{1}{100} = \frac{1}{400 }= 0.0025$

c) Calculate delta y when x =4 and delta x = 0.01. Interpret.

$\Delta y = f(x + \Delta x) - f(x)$

$\Delta y =\sqrt{4.01} - \sqrt{4} \approx 0.002498$

d) Illustrate dy and delta y. Interpret.

So what are they asking for "interpret" and "illustrate"?

Thanks,
Rod

2. ## Re: What is being asked?

Originally Posted by friesr
I have a 4 part problem that I am unsure what is being asked. I have shown what i have worked out.
$Lety = f(x) = \sqrt{x}$
a) Find the differential of y
$dy=\frac{1}{2\sqrt{x}} dx$

b) Calculate dy when x=4 and delta x = 0.01. Interpret.
$dy= \frac {1}{4} * \frac{1}{100} = \frac{1}{400 }= 0.0025$

c) Calculate delta y when x =4 and delta x = 0.01. Interpret.
$\Delta y = f(x + \Delta x) - f(x)$

$\Delta y =\sqrt{4.01} - \sqrt{4} \approx 0.002498$

d) Illustrate dy and delta y. Interpret.

So what are they asking for "interpret" and "illustrate"?

To be truthful, I not sure what the expected answer is.

But $\sqrt{4.01} - \sqrt{4} \approx 0.002498$ tell you that $\sqrt{4.01}\approx 2.002498$

3. ## Re: What is being asked?

Originally Posted by friesr
I have a 4 part problem that I am unsure what is being asked. I have shown what i have worked out.

$Lety = f(x) = \sqrt{x}$

a) Find the differential of y

$dy=\frac{1}{2\sqrt{x}} dx$

b) Calculate dy when x=4 and delta x = 0.01. Interpret.

$dy= \frac {1}{4} * \frac{1}{100} = \frac{1}{400 }= 0.0025$
You know that you can think of the derivative as the slope of the tangent line, don't you? If you start from $(4, 2)$ and x increases by .01 then the linear approximation to y(x) says that y will increase by 0.0025.

c) Calculate delta y when x =4 and delta x = 0.01. Interpret.

$\Delta y = f(x + \Delta x) - f(x)$

$\Delta y =\sqrt{4.01} - \sqrt{4} \approx 0.002498$
This is saying that if, starting at (4, 2), x increases by .01, then y actually changes by 0.002498.
Pretty close to the linear approximation, before, 0.0025 isn't it?

d) Illustrate dy and delta y. Interpret.
Draw a graph! Draw the graph of $y= \sqrt{x}$ and its tangent line at x= 4. Draw a horizontal line from (4, 2) to (4.01, 2). $\Delta y$ is the distance from (4.01, 2) up to the graph of $y= \sqrt{x}$ and dy is the distance from (4.01, 2) to the tangent line.

So what are they asking for "interpret" and "illustrate"?

Thanks,
Rod