pls help me doing the exercises

abdulwq

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Pls help me solve these exercises step by step.
I have missed the first sem and now i have cleared up basics and dont have enough time so if some can solve i can understand myself.Also i have done them but not sure if they are correct.
Thanks
 

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Pls help me solve these exercises step by step.
I have missed the first sem and now i have cleared up basics and dont have enough time so if some can solve i can understand myself.Also i have done them but not sure if they are correct.
Thanks
Since you have done these problems - show us your work - we can help you fix your mistakes (if there are any).

Please share your work with us.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (e.g. "are these correct?")
 
ok i am 27 and studing Automation Engg..

In first question...part (a) I put values of x and solved it.aNS::::1/5^1/2

b)Shall i just take the partial derivative of f(x1,x2) with respect to x1 or i dont get this one.

c) its the same as b.
 
ok i am 27 and studing Automation Engg..

In first question...part (a) I put values of x and solved it.aNS::::1/5^1/2

b)Shall i just take the partial derivative of f(x1,x2) with respect to x1 or i dont get this one.

c) its the same as b.
For part (b) set the numerator equal to zero.
For part (c) set x1 + 2x2 = √(x12 + x22) ...........and square both sides .................... tell us what you find
 
for b) why are we putting numerator=0.....i have taken the derivative of the equation wrt x1 but i dont know how shall i satisfy that f(x1,x2)=0.......if i put f'(x1)=0 still it does not satisfy....

c)i get 4x1x​2+3x2​2
 
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for b) why are we putting numerator=0 .....i have taken the derivative of the equation wrt x1 but i dont know how shall i satisfy that f(x1,x2)=0.......if i put f'(x1)=0 still it does not satisfy....

c)i get 4x1x​2+3x2​2
b) Read the problem statement - carefully. It has nothing to do with f'.
c) 4x1x​2+3x2​2 = 0 → x​2(4x1+ 3x2) = 0 → x2 = 0 or 4x1 + 3x2 = 0 for f(x1,x2) = 1
 
ok..:) i got it.

Q.2:: i am confusing myself here ...i know f(x)=f(x1,x2.....xn) so does the answer would be PD of x1 and x2.

2b) how would i take PD of f wrt y as there is not y term.

c) for gradient we just calculate PD of x1 and put value of x1 and x2.
 
Q.3....df/dl=2Pi(-1/2.L-3/2.C1/2)
we get -1/PiL3/2C1/2......now we put value of L and C to get the answer .( any mistake in it pls point)


3b)If L is increased by 2% how would i change the equation???
 
Q.3....df/dl=2Pi(-1/2.L-3/2.C1/2)
we get -1/PiL3/2C1/2......now we put value of L and C to get the answer .( any mistake in it pls point)


3b)If L is increased by 2% how would i change the equation???
\(\displaystyle \dfrac{\partial f}{\partial L} = \left( \dfrac{1}{2\pi}\right) \left( \dfrac{-1}{2}\right) \left( \dfrac{1}{L^{3/2}C^{1/2}}\right) \)

The two 2's do not cancel .. your result should have 4pi in the denominator, not pi.

Evaluate at given L and C. If L changes by some amount, the change in f will be \(\displaystyle \frac{\partial f}{\partial L} \times \Delta L \)
 
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ok i have done all the critical points problems but i am stuck at this.Tomorow is my exam so i am really looking for quick help understand it.
f(x1,x2)=x14+4x12x22 - 2x12+2x22 -1

so fx=4x13+8x1x22 -4x1
and fx2= 8x12x2+4x2
now i am having difficulty to factorize the equation and find critical values.Pls help me do it so i can master it.
 
ok i have done all the critical points problems but i am stuck at this.Tomorow is my exam so i am really looking for quick help understand it.
f(x1,x2)=x14+4x12x22 - 2x12+2x22 -1

so fx=4x13+8x1x22 -4x1
and fx2= 8x12x2+4x2
now i am having difficulty to factorize the equation and find critical values.Pls help me do it so i can master it.
We ALWAYS need to see your work to see what you have tried and where you are stuck.

You should factor the partial derivatives, and set the factors to 0 to find the critical values.

\(\displaystyle \dfrac{\partial f}{\partial x_1} = 4 x_1 (x_1^2 + 2 x_2^2 - 1) = 0 \) ......if x1=0, or on an ellipse.

\(\displaystyle \dfrac{\partial f}{\partial x_2} = 4 x_2 (2 x_1^2 + 1)= 0 \) ......if (and only if) x2=0.
 
ok so if we take fx2
4x2=0 and x22=1/2
if i put the second value x2 into fx1 then i am not getting it further.
 
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ok so if we take fx2
4x2=0 and x22=1/2
if i put the second value x2 into fx1 then i am not getting it further.
Please pay more attention to what you are saying. If 4x2=0, then it is necessarily true that x22=0 as well. ANY point with x2=0 will have 0 for the partial derivative sith respect to x2.

Now look at the loci of all points where the x-partial is 0. Set x2=0. How many points are there with both partials equal to zero?
 
its x12=-1/2 it came from fx​2
pls pardon my mistakes as they are mailnly typing.
now if we put value of x1 in equation 1 then after factoring i still dont get value of
x​2
 
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its x12=-1/2 it came from fx​2
pls pardon my mistakes as they are mailnly typing.
now if we put value of x1 in equation 1 then after factoring i still dont get value of
x​2
By "it came from fx​2," perhaps you are looking at how fx​2 may be zero. Did you notice that you have set x12 negative? That is impossible! (not real). That is why I said

The partial derivative with respect to x2 is zero IF AND ONLY IF x2=0

Set x2=0 in the formula for the partial with respect to x1.
For what values of x1 is that derivative zero?
 
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