## mlddmlnog Group Title Calculus. Help. :'( one year ago one year ago

1. mlddmlnog Group Title

for what non-negative valuof b is the line given by $y=-\frac{ 1 }{ 3 }x+b$ perpendicular to the curve y=x^3?

2. UnkleRhaukus Group Title

take the derivative (with respect to x) of both equations , what do you get?

3. mlddmlnog Group Title

i got.. y'=-1/3+b and y'=3x^2 ..

4. mlddmlnog Group Title

waht do you do next?

5. UnkleRhaukus Group Title

try the derivative of the first equation again ( you made a mistake)

6. mlddmlnog Group Title

hm... i can't seem to find what my mistake was.. :/ is it the +b part??? because I do not know what to do with that...

7. UnkleRhaukus Group Title

b is really bx^0

8. mlddmlnog Group Title

so then, it would just be y'=-1/3???

9. UnkleRhaukus Group Title

thats better

10. UnkleRhaukus Group Title

so you have found the first derivatives of the those lines, the first derivative is the slope of the line

11. UnkleRhaukus Group Title

the product of the slopes of perpendicular lines is -1

12. mlddmlnog Group Title

yes.. uhm.. give me a few min to think please :)

13. UnkleRhaukus Group Title

sure,

14. mlddmlnog Group Title

15. UnkleRhaukus Group Title

ok , so $m_1\times m_2=-1$ $m_1=-\frac13\qquad\qquad m_2=3x^2$ solve for x

16. mlddmlnog Group Title

x=+1 or -1 .

17. UnkleRhaukus Group Title

good

18. UnkleRhaukus Group Title

now set $y_1=y_2$ take the first case (x=-1) and find b

19. UnkleRhaukus Group Title

does that make sense?

20. mlddmlnog Group Title

yes it does :) sorry. open study was being so slow .but i get it now. thank you! :)

21. UnkleRhaukus Group Title

what did you you get for an answer/

22. mlddmlnog Group Title

no what do i do..?