Wednesday
June 19, 2013

Posts by Mishaka

Total # Posts: 106

Calculus (Normal Line, please check my work)
The slope of the line normal to the graph of 4 sin x + 9 cos y = 9 at the point (pi, 0) is: Derivative: 4cosx - 9siny(dy/dx) = 0 (dy/dx) = (-4cosx) / (-9siny) (dy/dx) = (4) / 0 Normal line = -1 / (4/0) Does this mean that the slope of the normal line is undefined, or did I do ...

Calculus
Find the point on the graph of y = x^2 + 1 that’s closest to the point 8, 1.5. Hint: Remember the distance formula.

Calculus (Optimization)
Nevermind, that 4.42 was a mistake and my very original answer of 1.105940354 was absolutely correct!!! This is the right answer, I know it!

Calculus (Optimization)
Okay, so does this change my original answer of approximately 1.64 to 4.42??? The 4.42 came from putting your new values in the quadratic equation.

Calculus (Optimization)
Now I'm lost, I don't get why you changed the signs.

Calculus (Optimization)
I rechecked and found that 3x^2-22x+28 has the correct signs. Knowing this equation and the values I found from the quadratic equation, would you say that the 1.639079157 term is correct? (The 2.69 square inches came from squaring the 1.639079157).

Calculus (Optimization)
I think that you might have gotten the equation wrong, I think that it should be: 3x^2 - 22x + 28. When I put this equation into the quadratic equation, I got 5.694254177 and 1.639079157. So the squares that need to be cut out should have an area of approximately 2.69 square i...

Calculus (Optimization, Still Need Help)
I just wanted to correct something for my equation, it should be: V = (14 - 2x)(8 - 3x)(x), which simplifies to V = 112x - 44x^2 - 4x^3. Take the derivative: V' = 112 - 88x - 12x^2 Now all I need are the roots, any help? I think I found one around 1.10594, possibly?

Calculus (Optimization)
A rectangular piece of cardboard, 8 inches by 14 inches, is used to make an open top box by cutting out a small square from each corner and bending up the sides. What size square should be cut from each corner for the box to have the maximum volume? So far I have: V = (14 - 2x...

Calculus
Pretty sure I figured it out, 4/27. I found this by simplifying: ((1/3pi (h - 2/3 h))(4/9 r^2)) / (1/3 pi r^2 h)

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