anonymous
  • anonymous
Looking for someone to check my answer: Find an equation of the tangent line through the given point. x 2 y 3 + 15y = 34x, (3, 2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\frac{ 34-2x }{ 2y+15 }\]
anonymous
  • anonymous
thats the answer i got.
anonymous
  • anonymous
well that's what i had for y'

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More answers

asnaseer
  • asnaseer
You first need to use implicit differentiation to find an expression for \(\displaystyle\frac{dy}{dx}\). Then substitute x=3 and y=2 into that expression to get the slope of the tangent line at that point. So then you will know the slope of the tangent line and you also know it passes through the point (3,2) - use this information to calculate the equation of the tangent line.
anonymous
  • anonymous
was my y' equation correct?
asnaseer
  • asnaseer
it doesn't look correct to me - can you please list your steps so that I can help spot where you may have made a mistake?
anonymous
  • anonymous
\[x ^{2}y ^{2}+15y=34x\] \[2x*2yy'+15y'=34\] \[y'(2y+15)=34-2x\] \[y'=\frac{ 34-2x }{ 2y+15 }\]
asnaseer
  • asnaseer
I thought you had \(y^3\) in the equation listed in your question?
anonymous
  • anonymous
lol so it is.
anonymous
  • anonymous
\[y'=\frac{ 34-2x }{ 3y ^{2}+15 }\]
anonymous
  • anonymous
did i have it right the second time (right above what you just wrote)
asnaseer
  • asnaseer
sorry I meant:\[\frac{d}{dx}(x^2y^3)=(x^2)\frac{d}{dx}(y^3)+y^3\frac{d}{dx}(x^2)\]
asnaseer
  • asnaseer
you haven't used the chain rule correctly
asnaseer
  • asnaseer
I mean "product rule"
asnaseer
  • asnaseer
have a look here: http://www.1728.org/chainrul.htm
anonymous
  • anonymous
\[x ^{2}*3y ^{2}y'+y ^{3}2x+15y'=34\]
asnaseer
  • asnaseer
that is correct :)
anonymous
  • anonymous
okay so now i isolate y'
asnaseer
  • asnaseer
exactly
anonymous
  • anonymous
\[y'=\frac{ 34-2xy ^{3} }{ x ^{2}3y ^{2}+15 }\]
asnaseer
  • asnaseer
yup - now follow the other steps that I had listed above.
asnaseer
  • asnaseer
NOTE: we don't usually write an expression as \(x^23y^2\) - it is better to write it as \(3x^2y^2\)
anonymous
  • anonymous
ok.
asnaseer
  • asnaseer
the general rule of thumb is to write in this order: 1. Constants first 2. Then letters in alphabetical order
anonymous
  • anonymous
okay so now i sub in my points right?
asnaseer
  • asnaseer
correct
anonymous
  • anonymous
so i had: -14/123
asnaseer
  • asnaseer
perfect! just a couple of more steps to go now :)
anonymous
  • anonymous
is the equation: \[y-2=-14/123x+42/123\]
asnaseer
  • asnaseer
yes - that looks correct. I wouldn't have separated the two constants here (the -2 and the 42/123)
asnaseer
  • asnaseer
you may also want to multiply both sides by 123 to remove the fractions from the final equation.
anonymous
  • anonymous
so how would you make it look?
asnaseer
  • asnaseer
ok, you got to this equation:\[y-2=-14/123x+42/123\]first add 2 to both sides to get:\[y=-14x/123 + 288/123\]then multiply both sides by 123 - what will you get then?
anonymous
  • anonymous
y123=-14x+288
asnaseer
  • asnaseer
correct - but again, remember to write constants first - so 123y instead of y123
anonymous
  • anonymous
that's not really how you write the equation of a line though.
asnaseer
  • asnaseer
so I would write the final equation as:\[123y=288-14x\]
asnaseer
  • asnaseer
it is still an equation of a line. you can write it in "standard form" as follows:\[14x+123y=288\]
asnaseer
  • asnaseer
maybe you are only used to seeing it in the form: \(y=mx +c\)
asnaseer
  • asnaseer
One last advice before I leave - you wrote one of the terms in your original equation as: -14/123x this can sometimes be confused for: \[-\frac{14}{123x}\]so it is usually better to write it as: -14x/123

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