30 Sep 08
1 edit
Originally posted by EladarAh!!! Yes of course you are correct! I was thinking in general terms. All this time I was asking: "Why does he think that the second derivative of a function must be 0?" and wasn't thinking in the particular function we have in mind.
If the question is to find f(6) or F'(6), then you are correct. If the question is to find f'(6), then I'm correct.
Remember that a is a constant. If F'(6) equals a constant, then F''(6) = 0.
F'(6)=a=f(6) and so f'(6)=F''(6)=a'=0.
Sorry for the confusion.
30 Sep 08
1 edit
It is only second derivatives of functions defined in this way that are zero, at least the second derivatives at specific numbers.
Best101,
As for the limit of x/(x-1) being 0/0, the common problem is that people do not reduce before replacing h with zero. Always reduce to cancel out the h in the denominator before substituting the 0.
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30 Sep 08
2 edits
Originally posted by EladarBeing the case that he was stuck on using the definition lim as h approaches 0 of [f(x+h) - f(6)]/h....I'm going with Adam on this, or the more simple solution without using any integrals...I mean, a sesoned veteran of Calculus is well aware of how to use that definition..Integrals don't show up until the end of calculus 1, and this seems like a more mid term calc 1 problem
It is only second derivatives of functions defined in this way that are zero, at least the second derivatives at specific numbers.
Best101,
As for the limit of x/(x-1) being 0/0, the common problem is that people do not reduce before replacing h with zero. Always reduce to cancel out the h in the denominator before substituting the 0.
01 Oct 08
2 edits
Originally posted by adam warlockThanks. This stuff makes Algebra look like nothing.
F(x+h)-F(x)=(x+h)/(x+h-1)-x/(x-1)=[(x+h)(x-1)-x(x+h-1)]/[(x+h-1)(x-1)]=
=(x^2-x+hx-h-x^2-xh+x)/[(x+h-1)(x-1)]=-h/[(x+h-1)(x-1)]=
Now [F(x+h)-F(x)]/h=-1/[(x+h-1)(x-1)]
And taking the limit as h go to 0 we are left with F'(x)=-1/(x-1)^2
@joe shmo: Btw I'm just started Ap Calculus about a month and a half ago in High School.
Edit: Is there a difference b/w F(x) and f(x)? I didn't think it mattered.
01 Oct 08
1 edit
Originally posted by Best101As Eladar said, yes.
Thanks. This stuff makes Algebra look like nothing.
@joe shmo: Btw I'm just started Ap Calculus about a month and a half ago in High School.
Edit: Is there a difference b/w F(x) and f(x)? I didn't think it mattered.
There are many different types of notation and it is important that you understand the notatin you are using. Some people use short cuts as well. Say, you are mutliplying two functions h(x) = f(x) * g(x). You might see someone write it out as h(x) = fg or even h = fg.
You may see differentials noted as df/dx instead of f'(x) as well.
Don't get too hung up on things having to be in a particular notation. It is important that you understand what the notation represents though.
03 Oct 08
I think I need to get out of Calculus, cause this stuff just doesn't make any sense.
Can anyone answer this one. Also do you think this is an appropiate test question for someone who just started Calculus about a month ago?
Q: Find the x coordinates of all the points on the curve x^1/2 + 1/2x^2, besides x=1, at which the slope of the tangent line equals 3/2.
You don't have to answer it if you don't want to because know how annoying it could be to type numbers on a computer.
03 Oct 08
Originally posted by Best101Just think about what the slop of a curve at a given point as to do with its derivative and this exercise is piece of cake.
I think I need to get out of Calculus, cause this stuff just doesn't make any sense.
Can anyone answer this one. Also do you think this is an appropiate test question for someone who just started Calculus about a month ago?
Q: Find the x coordinates of all the points on the curve x^1/2 + 1/2x^2, besides x=1, at which the slope of the tangent line equa ...[text shortened]... it if you don't want to because know how annoying it could be to type numbers on a computer.
Also do you think this is an appropiate test question for someone who just started Calculus about a month ago?
I think that you need to study a little bit harder
03 Oct 08
Originally posted by adam warlockIt's actually quite cleverly worded, I think. A good example problem, and slightly more difficult that the standard grinder.
Just think about what the slop of a curve at a given point as to do with its derivative and this exercise is piece of cake.
[b]Also do you think this is an appropiate test question for someone who just started Calculus about a month ago?
I think that you need to study a little bit harder[/b]
Since the slope is just the first derivative of the function, we start there:
y = x^(1/2) + (1/2)x^2
y' = (1/2)x^(-1/2) + x
Now we're looking for slopes equal to 3/2, so we let y' = 3/2:
3/2 = (1/2)x^(-1/2) + x
Multiplying by 2 to clear fractions, and grouping irrationals on one side:
2x - 3 = x^(-1/2)
Since the original function included a square root, we know that x is strictly greater than or equal to 0, so there's no problem with squaring both sides to clear the square root:
4x^2 - 12x + 9 = 1/x
Multiplying through by x and grouping all terms on one side:
4x^3 - 12x^2 + 9x - 1 = 0
Now we know that x = 1 is a solution to the above equation, because it's given in the problem statement. After factoring out (x - 1) using long division, we have:
(x - 1)(4x^2 - 8x + 1) = 0
So x = 1, or (4x^2 - 8x + 1) = 0. Solving this quadratic for x, we get:
x = (2 +/- SQRT(3))/2
Plugging these two solutions back into the original equation y' = (1/2)x^(-1/2) + x, we find that (2 - SQRT(3))/2 works, but (2 + SQRT(3))/2 doesn't. Therefore, the answer is (2 - SQRT(3))/2.
03 Oct 08
Originally posted by PBE6If my fist was travelling at mach 1.5 on an inverse tangental plane perpendicular to your face at a distance of 1 metre, what would be the expected kip force upon impact?
It's actually quite cleverly worded, I think. A good example problem, and slightly more difficult that the standard grinder.
Since the slope is just the first derivative of the function, we start there:
y = x^(1/2) + (1/2)x^2
y' = (1/2)x^(-1/2) + x
Now we're looking for slopes equal to 3/2, so we let y' = 3/2:
3/2 = (1/2)x^(-1/2) + x
Multiply ...[text shortened]... QRT(3))/2 works, but (2 + SQRT(3))/2 doesn't. Therefore, the answer is (2 - SQRT(3))/2.
03 Oct 08
Originally posted by uzlessCleverly worded! Assuming a mass of approximately 0.5 kg, a temperature of 15 C and sea-level pressure, the total momentum of your hand would be:
If my fist was travelling at mach 1.5 on an inverse tangental plane perpendicular to your face at a distance of 1 metre, what would be the expected kip force upon impact?
P = m*v = (0.5 kg)*(340.3 m/s x 1.5) = 255.2 kg*m/s
Also, assuming that your fist is uniformly decelerated into my face from its initial velocity to a dead stop over a period of 0.1 s, the total force would be:
F = dP/dt = (255.2 kg*m/s) / (0.1 s) = 2552 N
Since one pound-force is equal to 4.44822 N, one kip is equal to 4448.22 N. Converting, we have:
F = 2552 N * (1 kip / 4448.22 N) = 0.5738 kip = 0.6 kip
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