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**CIV**- Replies: 3

There's a right triangular pyramid with a lateral surface area of 195 root 3 and a slant height of 13. What's the volume?

That's the problem. I got 150 root 3 u^3 as an answer. A friend of mine got 459.86 u^3.

I saw my friends work. I understand some of it, but other parts not so much. Eventually I'll sit down, when I have more time, and figure out what she did.

I just want to get a third person, or more, to try it for an answer. Thanks.

I made a mistake.... darn it. Sorry.

It's:

It's continuous and I just figured out that it's not differentiable at 0. The slope at zero is undefined.

I always make mistakes like that when posting online. Usually when I post online my brain is hurting from trying to figure things out.

**CIV**- Replies: 3

The book I'm using says that the Mean Vaule Theorem does not apply with this function:

Why is this? It's continuous and differentiable right?

Using the MVT formula I get:

Then I solve for it's derivative and get:

Then I determine where the slope is 1/2:

I don't understand. The book says as the answer is "Does not apply".

**CIV**- Replies: 1

I know it to be the difference divided by the average and multiplied by 100. My one professor said to calculate the % difference you basically: 100((a - b) / b). I'm doing linear approximation in my math book and it wants the % difference and it mentioned the same thing: 100((approx - exact) / exact). What is this? I'm confused.

I meant 9 / root 13. I still don't follow what your doing there.

thickhead wrote:

Yes. I am getting

continuing with my previous post#8

How did you get 9 / root 5? I don't understand what you mentioned.

I just solved another optimization problem, part a of this problem, on my own without any help, except checking my calculations online. I didn't google the actual problem to see how its set up. Totally got it right and I'm so happy. Totally awesome:)

The problem: If you had a piece of wire 60 units in length, where would you cut this wire to form a square and a circle with A: the minimum combined area and B: the maximum combined area.

It involved making calculations with pi, it was confusing me a bit. Im not used to working with something like 2(pi)x as a term. Just need to solve for the max now.

The book has this as the answer:

This book doesn't usually have wrong answers so Im really trying to exhaust myself here to make sure it's not me.

Again, the book has 8 / root 5 as the answer to part one too. LOL. So I'm really scratching my head here.

bob bundy wrote:

hi CIV,

Looking again at your working, it seems we are using a different interpretation of part 2. 2 root13 is the distance across the sea from start to finish. So you aren't making any use of the walking speed. I thought the question meant row straight to the shore, as quickly as possible, and then walk 6 miles along the shore. I assumed this as part 2 repeats the walking speed. Why say this if no walking is involved?

Anyway, let's go back to part 1 first. Is the answer we both have considered to be correct by the book?

Bob

Yes. The answer for part 1 is correct. I corrected my typo.

Ok so yea I didn't type out part b word for word because I thought what I wrote was pretty clear, but that OK. Here's exactly what part b says:

**If she walks at 3 mi / hr, what is the minimum speed at which she must row so that the quickest way to the restaurant is to row directly (with no walking)?**

That's word for word. I probably should of said row directly to restaurant:/ Now you see what I mean about the book mentioning walking speed for part b? I know the distance to the restaurant directly by sea is 2 root 13 and I know what the quickest time is, so all I need is the rate at which to row to at least achieve the quickest time. That would be the minimum rate at which to row, which really isnt all that much faster than the original 2 mi / hr rowing speed.

bob bundy wrote:

Why are we even bothering with homework that is 10 years out of date?

Besides which, I'm not sure I like the PRONTO. THANX .... (1) Pronto seems a bit demanding; (2) Thanks isn't even spelt correctly .

Bob

Exactly. HA HA HA.

This is a question I had asked: http://www.mathisfunforum.com/viewtopic.php?id=23783

I made a serious effort to do the problem and made sure that everyone could see this. It is homework for me, but clearly you can see that I'm trying.

bobbym wrote:

I think he missed his deadline.

Probably.

You first. What have you done to try and solve this? Show us.

**CIV**- Replies: 16

It's the rowing and walking problem.

Basically, you're 4 miles out at sea from the closest point on shore and from that point on shore is a restaurant 6 miles away. You row at a speed of 2mi/hr and walk at a speed of 3mi/hr.

A: At what point on the shore should you land to minimize the travel time.

B: If you walk at a speed of 3mil/hr, what's the minimum speed at which you should row so that the quickest way is to row directly.

Now I already searched online for help and I got the answer to part A, which is:

I calculated the quickest travel time as well:

I compared the quickest time with the limits of the interval, that being 0 and 6. When x = 0, T = 4 hours. When x = 6, T is about 3.61 hours.

Now... the second part has me pulling my hair out because I feel like it's such an easy thing to figure out, but I'm not getting what the book has listed as the answer:

So, what I know is that you will be rowing a distance of:

I also know that the quickest travel time is:

So what I need now is the rate at which you must row.

If I use my answer to calculate total direct rowing time I get about 3.49 hours which is what it was before with some rowing and walking. This is the MINIMUM rowing rate. If I use the answer in the book I get 2.88 hours????

Am I missing some here? Is the answer in the book wrong? Did I not understand the question? I wrote part B here just about word for work just in case I misunderstood. I thought it was weird when it read "If you walk at a speed of 3mi/hr", because what does that have to do with the minimum rowing speed for rowing directly to be the fastest route.

Thanks everyone.

Zeeshan 01 wrote:

The only roots are 0 and 0. So (0,1) is a possible local extreme.

Why you not wote (0,1)

How y axis is 1

The location of the extrema is x = 0. The location of the tangent line, when it's slope is zero, is (0,1)

If you want the answers, you have to show us your best effort at trying to solve them first.

Identity wrote:

Each of us is of the form

, whereWe need to prove that b = 0

I definitely cross the x-axis at least once.....

How about you stay up all night and study math. Being amazing at math is way more important.

**CIV**- Replies: 4

I don't know why the book has me doing this.... I wasn't taught how to solve something like this yet. This is actually part of a calc 1 problem. Im graphing functions. It's part of the numerator for the second derivative. There's actually a 2 that I factored out, but it shouldn't be need to solve this. Well, technically now that I read the book it wants me to use a graphing calculator to find the important points. This polynomial is whats need to calculate possible inflection points.

I tried rational roots and it provided me with nothing that works, lol.

The derivative of sin x is cos x. Derivatives are equations used to calculate the SLOPE of a tangent line at some point on a curve. The curve is a function such as X^2 or in your example sin x. The derivative of X^2 is 2x. 2x is a linear equation and therefore is refereed to as a SLOPE equation. You can use that slope equation to calculate the ACTUAL slope of a tangent line on the curve of X^2 by choosing a point within the domain of the curve. Lets choose x = 2. The slope would be 2(2) = 4. The slope is represented with the letter m, so m = 4. Now that you have the slope, you need a y value to go with your x value of 2. Use that x value with the original equation to get yourself a y value: (2)^2 = 4. So now you have yourself a the slope of a tangent line at the point (2,4) on the curve X^2. Now what you need is the equation of this tangent line at that point. You remember the point-slope formula? Plug in all the information you have and solve for the equation for that tangent line: y - y2 = m(x-x2); y - 4 = 4(x - 2); y = 4(x - 2) + 4 = 4x - 8 + 4 = 4x - 4. So you have a point (2, 4); a slope equation of f'(x) = 2x; a slope of m = 4; and an equation of a tangent line, y = 4x - 4, at that point.

That's a basic derivative explanation. Forgive me if it appears that I'm assuming you know nothing.

It works the same for the function sin x. The slope equation sin x is f'(x) = cos x. One thing to keep in mind is that the graph of sin x goes to infinity to both direction. It repeats, goes round and round the unit circle. This is why your given an interval of (-pi, pi).

Now, it says determine when f is increasing and decreasing. A straight line is increasing when it's slope is positive and is decreasing when it's slope of negative. So if m = -2, that line is decreasing and if m = 4, that line is increasing. Positive increasing, negative decreasing.

So you have a function sin x and a slope equation for that function, which is cos x. Do you know you know how to find minima and maxima? The slope of minima and maxima is zero, m = 0. So now you know how to find minima and maxima. You set the first derivative, aka the slope equation for the function sin x, to zero. So now that cos x = 0, you have to ask yourself when is cos x zero? Cos x is zero at pi/2 and 3pi/2.

You might be asking why you have to find minima and maxima? By finding minima and maxima, you'll know where to check for increasing and decreasing slopes.

So now that you know that cos x = 0 at pi/2 and 3pi/2, are these values within your interval? pi/2 is, but 3pi/2 is not. If pi/2 is within the interval (-pi, pi), then so is -pi/2.

So you have these two points now. At these points cos x = 0 and being that cos x is the slope equation for sin x... the points on sin x with a slope of zero are the minima and maxima, the high and low points.

Now that you know where the high and low points are on sin x, those being -pi/2 and pi/2, you now know where to look for increasing and decreasing. You look between the high and low points, the minima and maxima. So you need a point between 0 and pi/2, and pi/2 and pi. Let's chose pi/4 and 3pi/4. If pi/4 and 3pi/4 are in the interval (-pi, pi) then so is -pi/4 and -3pi/4.

Now that we have points between -pi, -pi/2, pi/2, and pi; we can check for increasing and decreasing. Just use your calculator for this. f'(pi/4) = cos (pi/4) = 0.71, is that increasing or decreasing? Its increasing. Now we check f'(3pi/4) = cos (3pi/4) = -0.71, is that increasing or decreasing? It's decreasing. Do that rest and you we see whats happening.

Now remember! SIN X is the FUNCTION or CURVE. COS X is the DERIVATIVE or SLOPE EQUATION for a tangent line touching the function sin x. 0.71 and -0.71 are SLOPES of the tangent lines, the slope being the variable m, at the points pi/4 and 3pi/4.

If you want, you can solve for the equations of the tangent lines that same way I did in the very first explanation. Solve sin x using your x values to get y values. Using those y values, the x values, and the slopes, you can use the point-slope formula to solve for the tangent line equations.

I hope this helps. I can't believe I spent this much time writing all this out....

Thank you Bob. Yes, I mean maxima/ minima. The book I'm using refers to these points as critical points, sometimes extrema as well. I read through the section again and the book does say that there is extrema when f'(c) = 0 or f'(c) = dne. I guess being so tired I wasn't thinking straight. Also, after moving onto the next section it clearly states:

extrema when f'(c) = 0 or f'(c) = dne

and

possible inflection when f''(c) = 0 or f''(c) = dne

Thanks for your time bob.