Question

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

Answer 1

take the derivative and set it 0

Answer 2

im stuck on getting the deriviative

Answer 3

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

Answer 4

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

Answer 5

after found m use this formula and write the eq.

y-y1=m(x-x1)

y1=5

x1=4

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

use your cal..

Answer 10

-13/50 approx.

Answer 11

its showing as incorrect
what was the exact answer?

Answer 12

what are the options..

Answer 13

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

Answer 14

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

Answer 15

or \[y-4=(\frac{4\sqrt{3}-2\sqrt{10}}{2\sqrt{10}-5\sqrt{3}})(x-5)\]

Answer 16

is it saying round your answer or exact

Answer 17

exact

Answer 18

\[y=\frac{4\sqrt{3}-2\sqrt{10}}{2\sqrt{10}-5\sqrt{3}}x-5\frac{4\sqrt{3}-2\sqrt{10}}{2\sqrt{10}-5\sqrt{3}}+4\]

Answer 19

let me double check it

Answer 20

wht would the slope be?

Answer 21

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

Answer 22

slope is coeff of x

Answer 23

its sill coming up as incorrect

Answer 24

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

Answer 25

still coming up incorrect :/

Answer 26

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