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anonymous

  • 5 years ago

Find the equation of the tangent line to the curve y2+4xy+x2 = 13 at the point (2; 1).

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  1. bahrom7893
    • 5 years ago
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    find the first derivative: 2y*y' + 4(x*y' + y) + 2x = 0 2y*y' + 4x*y' + 4y + 2x = 0

  2. bahrom7893
    • 5 years ago
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    all you have to do now is find y' as it is also the slope of the tangent line. Plug in 2 for x and 1 for y: 2*y' + 4*2*y'+4*1+2*2 = 0 2y'+8y' = -4-4 10y' = -8 y' = M = -8/10

  3. bahrom7893
    • 5 years ago
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    Now use the formula: y - y1 = M(x-x1) y-1 = (-4/5)(x-2)

  4. bahrom7893
    • 5 years ago
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    M = -8/10 = -4/5 thats how i got -4/5 in the previous part..

  5. bahrom7893
    • 5 years ago
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    Simplify: y-1 = (-4/5)(x-2) y = -4x/5 + 8/5 +1 y = -4x/5 + 13/5

  6. bahrom7893
    • 5 years ago
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    Please click on become a fan if I helped, I really want to get to the next level!! Thanks =)

  7. anonymous
    • 5 years ago
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    yeah that helped a lot! it's been a few years since i did this stuff so i totally forgot about using y' as an intermediate variable, makes sense now tho, thanks and I will fan you :)

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