## tburn Group Title Can some help with question 2 and 4? I will post document. 2 years ago 2 years ago

1. Akshay_Budhkar Group Title

A] Propose a value for the following definite integral and give an interpretation for the result : is nothing but the area under the curve from the values x=6 to x=8

2. saifoo.khan Group Title

@Akshay_Budhkar , we can find that by simply using area of a square? right?

3. Akshay_Budhkar Group Title

yes 8^2=64 =D

4. saifoo.khan Group Title

Perfect. ;D

5. saifoo.khan Group Title

exactly..

6. Akshay_Budhkar Group Title

aha the y-axis is degrees / hour! not degrees!

7. saifoo.khan Group Title

Lol.

8. saifoo.khan Group Title

It's not easy. haha

9. Akshay_Budhkar Group Title

u already did :P i am trying

10. saifoo.khan Group Title

haha.

11. tburn Group Title

Amistre64 was helping me on this question but I was still lost so I was hoping if someone else also explained it maybe I would understand.

12. Akshay_Budhkar Group Title

ok i get it

13. Akshay_Budhkar Group Title

14. Akshay_Budhkar Group Title

are you clear with A ?

15. saifoo.khan Group Title

@Akshay_Budhkar idk but i feel like we have to find some derivatives?

16. tburn Group Title

Yes I understand what you said with A.

17. Akshay_Budhkar Group Title

@saifoo.khan yes

18. Akshay_Budhkar Group Title

ok now let degrees be d and time be t so on y axis we have d/t and x axis we have t Now till x= 1 it is a straight line, with a slope -8 and intercept -8 so the equation is y=mx+c that is y=-8x-8 that is d/t=-8t-8 d=-8t^2=8t

19. Akshay_Budhkar Group Title

*d=-8t^2-8

20. Akshay_Budhkar Group Title

Now for minima or maxima the first derivative of the function is o therefore${d(d) \over dt }=0=-16t-8 \therefore t= 1/2$

21. Akshay_Budhkar Group Title

Now to find if it is a minima or a maxima we take the second derivative. if it is negative it is a maxima if positive minima ( mostly i maybe mixing them ) so in this case it is -16 which is negative. therefore it is a maxima at 1/2

22. tburn Group Title

Thank you so much I understood that! Would you be able to help me with question 4 by chance?

23. Akshay_Budhkar Group Title

ok lemme see, is it one unit for each square right?

24. tburn Group Title

That is what I understand.

25. Akshay_Budhkar Group Title

ok lets start with the simplest C. f(3) = 5 any doubt?

26. tburn Group Title

Yes, thats what I see.

27. Akshay_Budhkar Group Title

ok now A. f'(-2) Now i believe that at a given point $f'(x)={dy \over dx} = { y \over x}$ if we estimate. so f'(-2) =-2/-2 = 1

28. tburn Group Title

Okay, that makes sense to me.

29. Akshay_Budhkar Group Title

for D its the same logic, area under the curve. i believe you can assume the graph to be a triangle and calculate the area

30. Akshay_Budhkar Group Title

is it unclear?

31. tburn Group Title

So you think D would be =-2 to 4?

32. Akshay_Budhkar Group Title

the area under the graph , from -2 to 4

33. tburn Group Title

Okay

34. Akshay_Budhkar Group Title

Can you solve it? For the other questions i would personally plot a graph for f'(x)

35. tburn Group Title

f"(x)=d/dx*f'(x)

36. Akshay_Budhkar Group Title

yes i know. Now to plot the graph u must know that it is zero at minima and maxima, and u can fidn approximate values by using the approximation of y/x, and plot the graph of f'

37. Akshay_Budhkar Group Title

then after i get the graph, f"' will be obtained by using y/x again and the last one by area under curve in that graph

38. tburn Group Title

Sorry I am just thinking. Okay, Thanks.

39. Akshay_Budhkar Group Title

yup give it a thought, and lemme know if ustuck =D and u can tag me in some other question if u get some similar doubt =D

40. tburn Group Title

Okay, thank you so much for your help!