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Math Help - Severation Questions that Need Explanations

  1. #1
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    Severation Questions that Need Explanations

    i got my test back, and i need to make corrections. parts of the questions i got only partial credit, while others i have no clue what to do.

    1. Part 1. If 3x^2+2xy+y^2=2, then the value of \frac{dy}{dx} at x=1 is: A. -2 B. 0 C. 2 D. 4

    I already figured out the derivative, which is y'=\frac{-y^2}{2xy-2}. Now I just have to solve for y. How do I do so?

    Part 2. Find all points on the curve where the slope is 0.

    Clueless on this part.

    Part 3. Write the equation of each vertical tangent line to the graph of f. (Hint: Vertical lines are x=#)

    I'm pretty sure this part is connected with part 2.

    2. How fast are the sides of a square changing at the instant when its sides are 6 feet long and its area is decreasing at a rate of 2 square feet per seconds?

    How would I find the rate? Do I take the derivative of the equation for area of a square? Some feedback would be nice.

    3. Let f and g be differentiable functions and let the values of f, g, and the derivatives f' and g' at x=1 and x=2 be given by the table below:
    x | f(x) | g(x) | f'(x) | g'(x)
    1 | 4 | 2 | 6 | 5
    2 | 3 | 1 | 7 | 8
    Determine the values of each of the following:
    a. The derivative of f+g at x=2
    b. The derivative of fg at x=2
    c.The derivative of \frac{f}{g} at x=2
    d. h'(2) where h(x)=f(g(x))

    I already got the answers for a-c (15, 56, 7/8 respectively), and I only need help with letter d.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by DarkestEvil View Post
    i got my test back, and i need to make corrections. parts of the questions i got only partial credit, while others i have no clue what to do.

    1. Part 1. If 3x^2+2xy+y^2=2, then the value of \frac{dy}{dx} at x=1 is: A. -2 B. 0 C. 2 D. 4

    I already figured out the derivative, which is y'=\frac{-y^2}{2xy-2}. Now I just have to solve for y. How do I do so?
    Use the initial equation.

    3(1)^2+2(1)y+y^2=2\implies y^2+2y+1=0 \implies (y+1)^2=0\implies y=-1

    Then just plug into the formula for y'. And by the way, I believe you calculated the derivative incorrectly. I should be \frac{dy}{dx}=-\frac{3x+y}{x+y}.

    Quote Originally Posted by DarkestEvil View Post
    Part 2. Find all points on the curve where the slope is 0.
    Just find where the numerator equals zero (and make sure the denominator doesn't equal zero at those points either).

    So the derivative is zero at all points (x,y) such that 3x+y=0 and 3x^2+2xy+y^2=2.

    Quote Originally Posted by DarkestEvil View Post
    Part 3. Write the equation of each vertical tangent line to the graph of f. (Hint: Vertical lines are x=#)

    I'm pretty sure this part is connected with part 2.
    Same deal as #2, except with the denominator. There will be a vertical tangent at all points (x,y) such that x+y=0 and 3x^2+2xy+y^2=2.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by DarkestEvil View Post
    2. How fast are the sides of a square changing at the instant when its sides are 6 feet long and its area is decreasing at a rate of 2 square feet per seconds?

    How would I find the rate? Do I take the derivative of the equation for area of a square? Some feedback would be nice.
    Some equations:

    A=s^2 so \frac{dA}{ds}=2s

    We want to find \frac{ds}{dt}, but we know that \frac{ds}{dt}=\frac{\frac{dA}{dt}}{\frac{dA}{ds}}

    Spoiler:
    We are given that \frac{dA}{dt}=2~ft^2/s. Since s=6, we can calculate that \frac{dA}{ds}=2\cdot6=12~ft. So,

    \frac{ds}{dt}=\frac{\frac{dA}{dt}}{\frac{dA}{ds}}=  \frac{2}{12}=\frac{1}{6}~ft/s
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by DarkestEvil View Post
    3. Let f and g be differentiable functions and let the values of f, g, and the derivatives f' and g' at x=1 and x=2 be given by the table below:
    x | f(x) | g(x) | f'(x) | g'(x)
    1 | 4 | 2 | 6 | 5
    2 | 3 | 1 | 7 | 8
    Determine the values of each of the following:
    a. The derivative of f+g at x=2
    b. The derivative of fg at x=2
    c.The derivative of \frac{f}{g} at x=2
    d. h'(2) where h(x)=f(g(x))

    I already got the answers for a-c (15, 56, 7/8 respectively), and I only need help with letter d.
    Just use the chain rule, and then plug in corresponding values from the table.

    Spoiler:
    By the chain rule, \frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x). Now just use the table: g(2)=1, f'(1)=6, g'(2)=8.

    So f'(g(2))\cdot g'(2)=f'(1)\cdot8=6\cdot8=48.
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  5. #5
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    Eh... I kinda made a mistake ^^"

    Messed up the first question. Its supposed to be:

    Consider the curve given by xy^2-2y=3
    a. Write a general expression for the slope of the curve.

    I got y'=\frac{-y^2}{2xy-2}

    b. Find all points on the curve where the slope is 0.

    I basically have to make the numerator of the fraction 0 right? I think I got that part down.

    c. Write the equation of each vertical tangent line to the graph of f.

    Same as part b, just with the denominator and the solutions should be x=something right?

    Sorry for the mix up lol.
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  6. #6
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    2xy-2 = 0

    2(xy - 1) = 0<br />

    xy = 1

    x= \frac{1}{y}


    sub \frac{1}{y} for x in the original equation for the curve ...

    \frac{1}{y} \cdot y^2 - 2y = 3

    y - 2y = 3

    y = -3

    vertical tangent equation ...

    x = -\frac{1}{3}
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  7. #7
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    Wow that makes so much more sense now, thanks for the help skeeter, you're always there for me lol
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  8. #8
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    Alright, I corrected most of the problems, but there are a few left I need help with.

    Using #3 from my first post, I was supposed to find:
    a. Derivative of f+g at x=2
    b. Derivative of fg at x=2
    c. Derivative of \frac{f}{g} at x=2

    Now the answer to the first one is correct, which was 15 [ f'+g'=(7)+(8)=15] I thought I had to do the same with the other ones, yet to my dismay, that wasn't the case. My teacher says to take the derivative before plugging in the values.


    Another problem I'm dealing with it:
    If 3x^3+2xy+y^2=2, then the value of \frac{dy}{dx} at x=1 is:
    A. -2 B. 0 C. 2 D. 4 E. not defined

    I already figured out the derivative is y'=\frac{-6x-2y}{2x+2y}
    My teacher says to find the value of of y first. So I used the first equation to find y=-1. I try plugging in both x and y into my derivative equation, then I get: y'=\frac{-6(1)-2(-1)}{2(1)+2(-1)} \Longrightarrow y'=\frac{-6+2}{2-2} which would give me \frac{-4}{0} which is undefined. So is the answer E?

    Thanks to everyone in advance.
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