exponential and logarithm problems (1 Viewer)

[xeriphic]

can someone help me out with these questions

1) the region under the graph y=1/2(e^x + e^-e) and between the lines x = -2 and x = 2 is rotated about the x-axis. Find the volume of the solid of revolution formed.

2) Find the second derivative of log(3x+1)

3) If y=logx, show that y" + (y')^2 = 0

4) The line x=a cuts the curves y=logx and y=log2x at P and Q. Show that the distance PQ remains constant for all values of a.

5) Show that the minimum value of y=xlogx is -1/e.

6) What the the graph of y=log3x look like, would it actually shift across the x axis or steeper than the logx graph, can some one explain

also important can someone tell me to what extent fo we have to know about curve sketching of exponential and logarithms,
i mean are we expected to know complex graphs involving them or just typical ones like y=1/2(e^x + e^-x), also can anyone suggest what would be the best way to plot these graphs, to actually go through first derivative and second derivative or just plot the points

is it possible to put the base of something else like 3 into the calculator apart from log10 or loge

thanks for the help, greatly appreciated

 

[:: ck ::]

Originally posted by xeriphic
can someone help me out with these questions

1) the region under the graph y=1/2(e^x + e^-e) and between the lines x = -2 and x = 2 is rotated about the x-axis. Find the volume of the solid of revolution formed.

wots the answer? i got : 22 units³ (to nearest units³... for y=(1/2)(E^x+E^-E)

probably wrong though ><"

if its rite ill post up working

 

[:: ck ::]

Originally posted by xeriphic
2) Find the second derivative of log(3x+1)

use the rule if y=ln[f(x)], y'=f'(x)/f(x)

d[ln(3x+1)]/dx = 3/(3x+1)
d²[ln(3x+1)]/dx² = -3(3)/(3x+1)²
= -9/(3x+1)²

 

[:: ck ::]

Originally posted by xeriphic

3) If y=logx, show that y" + (y')^2 = 0

y' = 1/x
y''= -1/x²

y'' + (y')²
= -1/x² + (1/x)²
= 0

 

[:: ck ::]

Originally posted by xeriphic
4) The line x=a cuts the curves y=logx and y=log2x at P and Q. Show that the distance PQ remains constant for all values of a.

y=logx
y = log2x

if x = a cuts these curves... we know the x coordinate of both curves will be a... and just sub into y to find the y coordinate

y=loga ... y = log2a

so coordinate of P is (a,loga) .. Q is (a,log2a)

distance formula...

PQ = sqrt( (a-a)^2 + (loga-log2a)^2)
= | loga-log2a | (distance must b +ve)
= |log(a/2a)|
= |log(1/2)| = log2

 

[:: ck ::]

Originally posted by xeriphic
5) Show that the minimum value of y=xlogx is -1/e.

this is straight forward... just get first deriv.. then use second deriv to show that it is greater than zero

y = xlnx
y' = x(1/x) + lnx
= 1+lnx

for stat pt... y'=0
0 = 1+lnx
lnx=-1
1/e = x

sub into y = xlnx
y = (1/e)ln(1/e)
= -1/e

but to show its a minimum u need to also show second derv is greater than zero for y = 1/e

y' = 1 + lnx
y''=1/x

sub x = 1/e
y'' = e
> 0

threfore at (1/e,-1/e) there is a min tp

therefore minimum value of y=xlnx is -1/e

 

[:: ck ::]

Originally posted by xeriphic
6) What the the graph of y=log3x look like, would it actually shift across the x axis or steeper than the logx graph, can some one explain

from one of the questions, the line x=a cutting log3x and logx will be a constant.. so its basically parallel to y=logx however the x intercept will be at (1/3,0)

 

[CrashOveride]

Originally posted by :: ryan.cck ::
wots the answer? i got : 22 units³ (to nearest units³... for y=(1/2)(E^x+E^-E)

probably wrong though ><"

if its rite ill post up working

I got 13.88pi units³

Ah wait, you're right its 22units³

 

[xeriphic]

1) the answer was 49.150 units cubed, i'm not sure how to do it still

5) i had the same working right up to 1/e, then i forgot to continue

6) did you get the x-intercept just by subbing 0 for y right, also can you tell about that program for drawing graphs it seems really helpful

i hate question 2 and 3, after i looked at it finally understood that i can just use the usual calculus to solve for second derivative

thanks for all your help, really appreciated

 

[:: ck ::]

Originally posted by xeriphic
1) the answer was 49.150 units cubed, i'm not sure how to do it still

can u check if u wrote the original question properly? i still get 22units cubed

6) did you get the x-intercept just by subbing 0 for y right, also can you tell about that program for drawing graphs it seems really helpful

yup so 0=log3x

rhs will b 0 when 3x = 1
x=1/3


i used winplot

 

[xeriphic]

ahh your right it should be y=1/2(e^x + e^-x)

just for reference can you also post how you did y=1/2(e^x + e^-e) thanks

 

[:: ck ::]

here ya go.. sorry abt handwriting.. kinda messy

click below for attachment

 

[CrashOveride]

Actually it is 22units cubed.

 

[xeriphic]

thanks both for taking the time especially for the scanning

just for question to find the y-coordinate is this right

y=(1/e)(ln1/e)

y= (lne^-1)/e

y= (-1lne)/e <--- question here is it okai to take down the -1

y= -1/e

 

[:: ck ::]

most likely a calculation error u had =P

it gets messy on calculator

 

[:: ck ::]

Originally posted by xeriphic
thanks both for taking the time

just for question to find the y-coordinate is this right

y=(1/e)(ln1/e)

y= (lne^-1)/e

y= (-1lne)/e <--- question here is it okai to take down the -1

y= -1/e

yes all acceptable.. remember lnaⁿ = nlna

 

[xeriphic]

Quote from before

also important can someone tell me to what extent fo we have to know about curve sketching of exponential and logarithms,
i mean are we expected to know complex graphs involving them or just typical ones like y=1/2(e^x + e^-x), also can anyone suggest what would be the best way to plot these graphs, to actually go through first derivative and second derivative or just plot the points

thanks

 

[:: ck ::]

ur expected to know these

y=+/- e^(+/-)x ...
and y = +/-Ln(x)

as for others such as y=1/2(e^x + e^-x)

just go thru first and second derivative... note the limits of the graph... any asymptotes.. x and y intercepts and when applicable, odd or even properties

if ur not satisfied or sure after... plot points and see if ur graph corresponds to the values u get on the calculator

 

[xeriphic]

i see now, thanks for all your help