cutie5507
Find the slope of the tangent line to the curve sqrt(3x +3y) +sqrt(2xy) =11.5
at points (4,5)



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cinar
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take the derivative and set it 0

cutie5507
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im stuck on getting the deriviative

cinar
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1/2(3x+3y)^(1/2)(3+3y')+1/2(2xy)^(1/2)(2y+2xy')=0

cinar
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sub x=4 and y=5 find y'=m

cinar
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after found m use this formula and write the eq.
yy1=m(xx1)
y1=5
x1=4

cinar
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\[\frac{1}{2}(3x+3y)^{1/2}(3+3y')+\frac{1}2 (2xy)^{1/2}(2y+2xy')=0\]

cutie5507
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i dont think im getting the right answer
what did you get as the answer?

cinar
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\[y'=m=\frac{4\sqrt{3}2\sqrt{10}}{2\sqrt{10}5\sqrt{3}}\]

cinar
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use your cal..

cinar
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13/50 approx.

cutie5507
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its showing as incorrect
what was the exact answer?

cinar
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what are the options..

cutie5507
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i dont have options, but the assignment is online, so it tells me if he answer is correct or not

cinar
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\[y=\frac{13}{50}x+\frac{53}{10}\]

cinar
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or \[y4=(\frac{4\sqrt{3}2\sqrt{10}}{2\sqrt{10}5\sqrt{3}})(x5)\]

cinar
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is it saying round your answer or exact

cutie5507
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exact

cinar
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\[y=\frac{4\sqrt{3}2\sqrt{10}}{2\sqrt{10}5\sqrt{3}}x5\frac{4\sqrt{3}2\sqrt{10}}{2\sqrt{10}5\sqrt{3}}+4\]

cinar
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let me double check it

cutie5507
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wht would the slope be?

cinar
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\[y=\frac{15\sqrt{2}+2\sqrt{15}}{12 \sqrt{2}+2\sqrt{15}}x+5\frac{15\sqrt{2}+2\sqrt{15}}{12 \sqrt{2}+2\sqrt{15}}+4\]

cinar
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slope is coeff of x

cutie5507
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its sill coming up as incorrect

cinar
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\[y=\frac{\sqrt{10}+4\sqrt{3}}{\sqrt{10}+4\sqrt{3}}x+5\frac{\sqrt{10}+4\sqrt{3}}{\sqrt{10}+4\sqrt{3}}+4\]

cutie5507
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still coming up incorrect :/

cinar
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\[y=\frac{12\sqrt{3}+15}{2\sqrt{3}+12}x+\frac{60\sqrt{3}75}{2\sqrt{3}+12}+4\]