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

1. cinar

take the derivative and set it 0

2. cutie5507

im stuck on getting the deriviative

3. cinar

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

4. cinar

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

5. cinar

after found m use this formula and write the eq. y-y1=m(x-x1) y1=5 x1=4

6. cinar

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

7. cutie5507

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

8. cinar

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

9. cinar

10. cinar

-13/50 approx.

11. cutie5507

its showing as incorrect what was the exact answer?

12. cinar

what are the options..

13. cutie5507

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

14. cinar

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

15. cinar

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

16. cinar

17. cutie5507

exact

18. cinar

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

19. cinar

let me double check it

20. cutie5507

wht would the slope be?

21. cinar

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

22. cinar

slope is coeff of x

23. cutie5507

its sill coming up as incorrect

24. cinar

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

25. cutie5507

still coming up incorrect :/

26. cinar

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