## mlddmlnog 2 years ago Calculus. Help. :'(

1. mlddmlnog

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

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

3. mlddmlnog

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

4. mlddmlnog

waht do you do next?

5. UnkleRhaukus

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

6. mlddmlnog

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

b is really bx^0

8. mlddmlnog

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

9. UnkleRhaukus

thats better

10. UnkleRhaukus

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

11. UnkleRhaukus

the product of the slopes of perpendicular lines is -1

12. mlddmlnog

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

13. UnkleRhaukus

sure,

14. mlddmlnog

15. UnkleRhaukus

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

16. mlddmlnog

x=+1 or -1 .

17. UnkleRhaukus

good

18. UnkleRhaukus

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

19. UnkleRhaukus

does that make sense?

20. mlddmlnog

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

21. UnkleRhaukus

what did you you get for an answer/

22. mlddmlnog

no what do i do..?