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Math Help - coordinates of P - parametric equations

  1. #1
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    coordinates of P - parametric equations

    Hi

    The parametric equations are x= e^t -1 and y=2e^t +3t

    So dx/dt = e^t and dy/dt = 2e^t + 3

    Let P be a point of curve C. Suppose the tangent line to C at the point P has slope 7. Find the coordinates of P

    dy/dx = (2e^t + 3)/e^t = 7 so 7= 2+ 3/e^t and e^t = 3/5

    Now I can then figure out that x = -2/5 but how do I figure out what y is?

    y = 2e^t + 3t => y = 2(3/5) + 3t

    Help please!

    Thanks

    Calculus beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    Hi

    The parametric equations are x= e^t -1 and y=2e^t +3t

    So dx/dt = e^t and dy/dt = 2e^t + 3

    Let P be a point of curve C. Suppose the tangent line to C at the point P has slope 7. Find the coordinates of P

    dy/dx = (2e^t + 3)/e^t = 7 so 7= 2+ 3/e^t and e^t = 3/5

    Now I can then figure out that x = -2/5 but how do I figure out what y is?

    y = 2e^t + 3t => y = 2(3/5) + 3t

    Help please!

    Thanks

    Calculus beginner
    t = \ln\left(\frac{3}{5}\right)
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  3. #3
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    So y = 6/5 + 3 ln (3/5) but what does that equal to make it a coordinate with x= -2/5?

    How do you add 6/5 + 3ln(3/5)?

    Thanks
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  4. #4
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    Quote Originally Posted by calcbeg View Post
    So y = 6/5 + 3 ln (3/5) but what does that equal to make it a coordinate with x= -2/5?

    How do you add 6/5 + 3ln(3/5)?

    Thanks
    you don't ... leave it as is
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