anonymous
  • anonymous
Can some help with question 2 and 4? I will post document.
Mathematics
schrodinger
  • schrodinger
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Akshay_Budhkar
  • Akshay_Budhkar
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
saifoo.khan
  • saifoo.khan
@Akshay_Budhkar , we can find that by simply using area of a square? right?
Akshay_Budhkar
  • Akshay_Budhkar
yes 8^2=64 =D

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saifoo.khan
  • saifoo.khan
Perfect. ;D
saifoo.khan
  • saifoo.khan
exactly..
Akshay_Budhkar
  • Akshay_Budhkar
aha the y-axis is degrees / hour! not degrees!
saifoo.khan
  • saifoo.khan
Lol.
saifoo.khan
  • saifoo.khan
It's not easy. haha
Akshay_Budhkar
  • Akshay_Budhkar
u already did :P i am trying
saifoo.khan
  • saifoo.khan
haha.
anonymous
  • anonymous
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.
Akshay_Budhkar
  • Akshay_Budhkar
ok i get it
Akshay_Budhkar
  • Akshay_Budhkar
lets start with A
Akshay_Budhkar
  • Akshay_Budhkar
are you clear with A ?
saifoo.khan
  • saifoo.khan
@Akshay_Budhkar idk but i feel like we have to find some derivatives?
anonymous
  • anonymous
Yes I understand what you said with A.
Akshay_Budhkar
  • Akshay_Budhkar
@saifoo.khan yes
Akshay_Budhkar
  • Akshay_Budhkar
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
Akshay_Budhkar
  • Akshay_Budhkar
*d=-8t^2-8
Akshay_Budhkar
  • Akshay_Budhkar
Now for minima or maxima the first derivative of the function is o therefore\[{d(d) \over dt }=0=-16t-8 \therefore t= 1/2\]
Akshay_Budhkar
  • Akshay_Budhkar
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
anonymous
  • anonymous
Thank you so much I understood that! Would you be able to help me with question 4 by chance?
Akshay_Budhkar
  • Akshay_Budhkar
ok lemme see, is it one unit for each square right?
anonymous
  • anonymous
That is what I understand.
Akshay_Budhkar
  • Akshay_Budhkar
ok lets start with the simplest C. f(3) = 5 any doubt?
anonymous
  • anonymous
Yes, thats what I see.
Akshay_Budhkar
  • Akshay_Budhkar
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
anonymous
  • anonymous
Okay, that makes sense to me.
Akshay_Budhkar
  • Akshay_Budhkar
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
Akshay_Budhkar
  • Akshay_Budhkar
is it unclear?
anonymous
  • anonymous
So you think D would be =-2 to 4?
Akshay_Budhkar
  • Akshay_Budhkar
the area under the graph , from -2 to 4
anonymous
  • anonymous
Okay
Akshay_Budhkar
  • Akshay_Budhkar
Can you solve it? For the other questions i would personally plot a graph for f'(x)
anonymous
  • anonymous
f"(x)=d/dx*f'(x)
Akshay_Budhkar
  • Akshay_Budhkar
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'
Akshay_Budhkar
  • Akshay_Budhkar
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
anonymous
  • anonymous
Sorry I am just thinking. Okay, Thanks.
Akshay_Budhkar
  • Akshay_Budhkar
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
anonymous
  • anonymous
Okay, thank you so much for your help!

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