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p3 parametrics watch

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    hey, stuck on a question :

    x = a sec t
    y = a tan t, and i need to find dy/dx

    sounds easy, but i dont know where to start from, to differentiate x and y, then apply the chain rule.

    anyone help plz?
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    (Original post by Taslima)
    hey, stuck on a question :

    x = a sec t
    y = a tan t, and i need to find dy/dx

    sounds easy, but i dont know where to start from, to differentiate x and y, then apply the chain rule.

    anyone help plz?
    cosect.
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    (Original post by Ralfskini)
    cosect.
    welldone, how did yu get it?
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    • Find dx/dt (put secx as 1/cosx and use quotient rule)

    • Find dy/dt

    • Then dy/dx = (dy/dt)/(dx/dt)


    The answer you should get is cosec(t)
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    basically have to use the rule:

    dy/dx = (dy/dt)/(dx/dt)

    y=atant
    dy/dt = a (sect)^2

    x= asect
    dx/dt = asect tant

    dy/dx = (asec^2t)/(asecttant)
    = sect/tant
    =1/(cost.tant)
    =1/sint
    =cosect

    Hope that helps
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    (Original post by el GaZZa)
    • Find dx/dt (put secx as 1/cosx and use quotient rule)

    • Find dy/dt

    • Then dy/dx = (dy/dt)/(dx/dt)


    The answer you should get is cosec(t)
    this doesnt help me at all.. what happens to the a's? in both x and y, and how do i "find" y = a tan t, its the whole reason why im posting, so i can be told "how" to do it..
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    (Original post by Taslima)
    this doesnt help me at all.. what happens to the a's? in both x and y, and how do i "find" y = a tan t, its the whole reason why im posting, so i can be told "how" to do it..
    a's cancel out when you divide one by the other
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    (Original post by Silly Sally)
    basically have to use the rule:

    dy/dx = (dy/dt)/(dx/dt)

    y=atant
    dy/dt = a (sect)^2

    x= asect
    dx/dt = asect tant

    dy/dx = (asec^2t)/(asecttant)
    = sect/tant
    =1/(cost.tant)
    =1/sint
    =cosect

    Hope that helps
    wow! thanks very very much!
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    hey, i have another

    x = t cos t
    y = t sin t

    find dy/dx, any helpers plz?
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    I posted it like that as an outline to parametric differentiation (essentially the rule that dy/dx = (dy/dt)/(dx/dt). The rest is standard differentiation (and I thought you were asking for help with the parametric part).

    Ok, well firstly, a is a constant, which is multiplied by the rest of the function. Because the function is being differentiated with respect to t, a is just a number and nothing happens to it. For example, d(3x^2)/dx = 3*(2x). This is exactly the same as if we differentiated x^2 and multiplied by 3 afterwards. So if you like, ignore the a while differentiating, and then multiply the answer by a afterwards.

    To find d(tanx)/dx, split tanx into sinx/cosx and apply the quotient rule. (Let u=sinx, v=cosx and plug into [v(du/dx) - u(dv/dx)]/(v^2)] ). Then do the same for secx (which is 1/cosx).

    It will benefit anyone's calculus skills to be able to do this for all 6 trigonometric functions, I suggest you take a few minutes to do this as it will give you a sounder base for calculus questions in the exam.

    Incidentally, the differential of tanx is used often and so its a good idea to learn it without having to work it out, again in case it comes up in exams.

    Let me know what answers you get
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    This time you've got to use the product rule [v(du/dx) + u(dv/dx)].

    (btw, I'm not trying to skimp on the workload or anything, but it's more beneficial to you if you were to do it!) Let me know what you get.
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    (Original post by el GaZZa)
    I posted it like that as an outline to parametric differentiation (essentially the rule that dy/dx = (dy/dt)/(dx/dt). The rest is standard differentiation (and I thought you were asking for help with the parametric part).

    Ok, well firstly, a is a constant, which is multiplied by the rest of the function. Because the function is being differentiated with respect to t, a is just a number and nothing happens to it. For example, d(3x^2)/dx = 3*(2x). This is exactly the same as if we differentiated x^2 and multiplied by 3 afterwards. So if you like, ignore the a while differentiating, and then multiply the answer by a afterwards.

    To find d(tanx)/dx, split tanx into sinx/cosx and apply the quotient rule. (Let u=sinx, v=cosx and plug into [v(du/dx) - u(dv/dx)]/(v^2)] ). Then do the same for secx (which is 1/cosx).

    It will benefit anyone's calculus skills to be able to do this for all 6 trigonometric functions, I suggest you take a few minutes to do this as it will give you a sounder base for calculus questions in the exam.

    Incidentally, the differential of tanx is used often and so its a good idea to learn it without having to work it out, again in case it comes up in exams.

    Let me know what answers you get
    aww! how sweet!! thank00 , that was very nice of yu, thankyu alot! i didnt intend on sounding mean hehe, but i got to sec t/tan t, on my own, until silly sally showed me the rest of it but, thankyu alot! , but ive learnt that if there's a constant infront, it is just normal differentiation and ignore the a, but if there's a variable, then it's product rule
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    Glad I could be of use
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    (Original post by el GaZZa)
    Glad I could be of use
    weeee!! just done it

    i gots me : (tcost + sint)/(cost - tsint)

    i hope thats right!
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    (Original post by Taslima)
    weeee!! just done it

    i gots me : (tcost + sint)/(cost - tsint)

    i hope thats right!
    Is the right answer!
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    (Original post by el GaZZa)
    Is the right answer!
    w00000!! go tazi!
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    okie, i have another question!! :P

    i need to find the equation of the normal to the curve with parametric equations
    x = 3t - t^3
    y = t^2
    at the point where t = 2

    this is what i have so far :

    dx/dt = 3 - 3t^2
    dy/dt = 2t

    => dy/dx = 2t/(3 - 3t^2)
    at [ t = 2 ], gradient is -4/9, therefore the normal gradient will be 9/4.

    now for the normal equation..

    y - y1 = m (x - x1)
    y - _ = 9/4 (x - _)

    how do i know what the x and y values are to put into the equation?
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    (Original post by Taslima)
    okie, i have another question!! :P

    i need to find the equation of the normal to the curve with parametric equations
    x = 3t - t^3
    y = t^2
    at the point where t = 2

    this is what i have so far :

    dx/dt = 3 - 3t^2
    dy/dt = 2t

    => dy/dx = 2t/(3 - 3t^2)
    at [ t = 2 ], gradient is -4/9, therefore the normal gradient will be 9/4.

    now for the normal equation..

    y - y1 = m (x - x1)
    y - _ = 9/4 (x - _)

    how do i know what the x and y values are to put into the equation?
    use the t value that you`ve used into the original parametric equation
    in your case its t = 2
    then
    x = 3(2) - (2)^3
    y = 2^2

    x = -2
    y = 4

    use your gradient and get the equation
 
 
 
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