tburn
Can some help with question 2 and 4? I will post document.
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Akshay_Budhkar
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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
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@Akshay_Budhkar , we can find that by simply using area of a square? right?
Akshay_Budhkar
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yes 8^2=64 =D
saifoo.khan
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Perfect. ;D
saifoo.khan
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exactly..
Akshay_Budhkar
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aha the y-axis is degrees / hour! not degrees!
saifoo.khan
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Lol.
saifoo.khan
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It's not easy. haha
Akshay_Budhkar
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u already did :P i am trying
saifoo.khan
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haha.
tburn
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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
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ok i get it
Akshay_Budhkar
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lets start with A
Akshay_Budhkar
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are you clear with A ?
saifoo.khan
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@Akshay_Budhkar idk but i feel like we have to find some derivatives?
tburn
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Yes I understand what you said with A.
Akshay_Budhkar
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@saifoo.khan yes
Akshay_Budhkar
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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
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*d=-8t^2-8
Akshay_Budhkar
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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
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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
tburn
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Thank you so much I understood that! Would you be able to help me with question 4 by chance?
Akshay_Budhkar
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ok lemme see, is it one unit for each square right?
tburn
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That is what I understand.
Akshay_Budhkar
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ok lets start with the simplest
C. f(3) = 5 any doubt?
tburn
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Yes, thats what I see.
Akshay_Budhkar
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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
tburn
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Okay, that makes sense to me.
Akshay_Budhkar
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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
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is it unclear?
tburn
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So you think D would be =-2 to 4?
Akshay_Budhkar
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the area under the graph , from -2 to 4
tburn
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Okay
Akshay_Budhkar
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Can you solve it?
For the other questions i would personally plot a graph for f'(x)
tburn
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f"(x)=d/dx*f'(x)
Akshay_Budhkar
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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
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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
tburn
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Sorry I am just thinking. Okay, Thanks.
Akshay_Budhkar
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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
tburn
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Okay, thank you so much for your help!