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October 30th, 2017, 09:55 PM   #1
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help please

i've spent hours on this problem from my calc i class and am stumped, can't seem to find the right approach to this besides guess and checking...if anyone could describe some steps to take since apparently it requires computer software to calculate, i would be so so grateful



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October 30th, 2017, 10:29 PM   #2
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$f_p(x) = 3 - \dfrac{x^2}{10}$
$f^\prime_p(x) = -\dfrac{x}{5}$

$f_c(x) = \sqrt{100-x^2},~x\in [-10,10]$
$f^\prime_c(x) = -\dfrac{x}{\sqrt{100-x^2}}$

Now solve for

$f^\prime_p(x) = f^\prime_c(x)$

If you can use mathematica it's trivial. By hand it's not so bad.
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October 30th, 2017, 11:30 PM   #3
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hey, thanks for the reply, but why do you decide to place the half circle above the parabola? i thought about this question in the way you presented, but i interpreted it where there is a single tangent line that is shared between a point on the parabola and another on the half circle. i'll ask my prof about this
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October 30th, 2017, 11:53 PM   #4
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Quote:
Originally Posted by ezmoney View Post
hey, thanks for the reply, but why do you decide to place the half circle above the parabola? i thought about this question in the way you presented, but i interpreted it where there is a single tangent line that is shared between a point on the parabola and another on the half circle. i'll ask my prof about this
The problem states the upper half of the circle.

Last edited by romsek; October 30th, 2017 at 11:56 PM.
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October 31st, 2017, 12:15 AM   #5
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Quote:
Originally Posted by romsek View Post
The problem states the upper half of the circle.
doesn't the problem ask for the upper half circle to be centered at (20, 0) not at (0, 0) provided with $\sqrt{100-x^2}$
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October 31st, 2017, 12:51 AM   #6
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Quote:
Originally Posted by ezmoney View Post
doesn't the problem ask for the upper half circle to be centered at (20, 0) not at (0, 0) provided with $\sqrt{100-x^2}$
sigh, yes it does. I missed that bit.

$f_c(x) = \sqrt{100 - (x-20)^2}$
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