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    How would I go about answering these 2 questions?
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    Hi TheEd, I see if I can be of some help

    5. Now I've never actually done a question like 5 before but I think I can see a potential way to solve it. So imagine the 3D polt of the surface

    z = ye^x

    where z is the vertical axis and y and x are horizontal axes. Now I read the question as looking for a horizontal direction in which the gradient of the surface is 0, so this means that we will be looking at a derivative where y and x are restricted, such that we have y = ax + b, where a and b are to be found. Now using that as a substitution for y we can find dz/dx and using the point given as well that should be enough info to solve for a and b, then once you have them you can determine a vector direction if thats what you require . I might be wrong, hopefully not though.

    6. Now as for the deriviative of
    y = 2e^{-x} \ \ \ at \ \ \ x = 0

    certainly you should simply be able to differentiate the function, and determine the gradient at x = 0, and you should be familiar with the equation:

    y - y_1 = m(x - x_1)

    using that you can determine the equation of the tangent line at x = 0. If you follow that through you should definatly see the connection to question 5, which makes me thing there is a more correct way to to 5 where the connection is not so obvious , hope thats help TheEd
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    (Original post by Galadirith)
    Hi TheEd, I see if I can be of some help

    5. Now I've never actually done a question like 5 before but I think I can see a potential way to solve it. So imagine the 3D polt of the surface

    z = ye^x

    where z is the vertical axis and y and x are horizontal axes. Now I read the question as looking for a horizontal direction in which the gradient of the surface is 0, so this means that we will be looking at a derivative where y and x are restricted, such that we have y = ax + b, where a and b are to be found. Now using that as a substitution for y we can find dz/dx and using the point given as well that should be enough info to solve for a and b, then once you have them you can determine a vector direction if thats what you require . I might be wrong, hopefully not though.

    6. Now as for the deriviative of
    y = 2e^{-x} \ \ \ at \ \ \ x = 0

    certainly you should simply be able to differentiate the function, and determine the gradient at x = 0, and you should be familiar with the equation:

    y - y_1 = m(x - x_1)

    using that you can determine the equation of the tangent line at x = 0. If you follow that through you should definatly see the connection to question 5, which makes me thing there is a more correct way to to 5 where the connection is not so obvious , hope thats help TheEd
    Thanks very helpful.

    I've solved for a=-2 and b=2 and clearly the 2 equations in 5 and 6 are y=-2x+2.

    But now how would I obtain a vector direction for question 5 and for 6 how would I explain clearly in words what the link between the 2 are?
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    Glad it helped .

    Well have a real think about what the gradient of any (linear) line represents "along the coridor and up the stairs" etc, literally if a gradient is 'm', and I move 'a' units in the x direction, how many units will I move in the y direction? What is a vector? have a think about that dont want to give any more away ;D.

    As for explaning the link, Im not really sure how much is nessesary, I assume this is a first year Maths Undergrad? I mean I feel it would be sufficient to simply state that the tangent line to the second equation at the point is the same as the tangent line of grad 0 blah blah blah. I think because we arrived at the equation y=-2x + 2 in question 5 the link is quite trivial so no indepth explaination is nessesary, but im sure there would be another way to solve 5, where one would produce a vector direction directly, as apposed to the method we employed getting our equation. I cant really think of any other real significance, its all quite right in front of your

    Hope thats helps TheEd
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    (Original post by Galadirith)
    Glad it helped .

    Well have a real think about what the gradient of any (linear) line represents "along the coridor and up the stairs" etc, literally if a gradient is 'm', and I move 'a' units in the x direction, how many units will I move in the y direction? What is a vector? have a think about that dont want to give any more away ;D.

    As for explaning the link, Im not really sure how much is nessesary, I assume this is a first year Maths Undergrad? I mean I feel it would be sufficient to simply state that the tangent line to the second equation at the point is the same as the tangent line of grad 0 blah blah blah. I think because we arrived at the equation y=-2x + 2 in question 5 the link is quite trivial so no indepth explaination is nessesary, but im sure there would be another way to solve 5, where one would produce a vector direction directly, as apposed to the method we employed getting our equation. I cant really think of any other real significance, its all quite right in front of your

    Hope thats helps TheEd
    A vector equation would be the following wouldn't it?

    \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 1 \\0 \end{bmatrix}+\lambda\begin{bmat  rix} -1 \\2 \end{bmatrix}
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    Yeh thats looks great to me . I was thinking simply the vector without the paremetric coefiecient would surfice for direction but your equation obviously gives the complete equation of the line in a paremetric vector form. Now there is one thing I would amened to your equation, we have discused this all relating to 2 'horizonatal' dimensions with z sticking up vertically that we havnt really thought about since the begining.

    So to properly get a full vector equation for x, y and z it would be:

    \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] =

\left[ \begin{array}{c} 0 \\ 2 \\ ? \end{array} \right] + \lambda \left[ \begin{array}{c} -1 \\ 2 \\ ? \end{array} \right]

    I had initialy filled in the ? marks then I thought, mmm shall I be a little more devious and make TheEd do the work HAHA . Note that I have changed the initial location to (x,y) = (0,2) , not thats not nessesary, but you would run into trouble if you simply took your (1,0) and did something with that :O, if you wanted me to elaborate on that a little more id be happy to if you solve the one above . Its up to you if you wanted to do that, Im not entirely sure what your required to do. That looks great though what youve done, its up to you to determine exactly howyou formaly write those answers now nice one.
 
 
 

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