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xvdavis

  • 5 years ago

Hello, I am preparing for my Fundamentals of Engineering Exam. Here is math problem that I have below which I am having difficulty with: Given: dy(1)/dx = 2/13 (1 + 5/2x - 3/2 - 3/4k) What is the value of k such that y(1) is perpendicular to the curve y(2)=2x at x=1? I have the solution to this problem if you need to see it (Which I don't understand.) Thanks!

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  1. mathteacher1729
    • 5 years ago
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    The notation you're using is a bit difficult to read. There is an "equation" button on the lower left hand corner of the text box which might make things easier. Here's my attempt to re-write your problem: \[\frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x-\frac{3}{2}-\frac{3}{4}k\] Find \[k\] such that \[y_1\] is perpendicular to \[y_2=2x\] when \[x = 1\] ?

  2. mathteacher1729
    • 5 years ago
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    Sorry, should be \frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x-\frac{3}{2}-\frac{3}{4}k) I forgot to close the parenthesis.

  3. mathteacher1729
    • 5 years ago
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    \[\frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x-\frac{3}{2}-\frac{3}{4}k\]

  4. mathteacher1729
    • 5 years ago
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    Grrrrr! \[\frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x-\frac{3}{2}-\frac{3}{4}k)\]

  5. anonymous
    • 5 years ago
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    let value of dy1/dx at x=1 be c. c x dy2/dx =-1 2c=-1 solve the equation

  6. anonymous
    • 5 years ago
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    dy1/dx at x=1 means substitute value of x as 1 in dy1/dx

  7. anonymous
    • 5 years ago
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    as the two curves are perpendicular at the givn point the product of their slopes should be -1

  8. xvdavis
    • 5 years ago
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    Thank you.

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