Calculus Calculus Math Forum

 October 30th, 2017, 08:55 PM #1 Newbie   Joined: Oct 2017 From: us Posts: 3 Thanks: 0 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
 October 30th, 2017, 09:29 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,120 Thanks: 1101 $f_p(x) = 3 - \dfrac{x^2}{10}$ $f^\prime_p(x) = -\dfrac{x}{5}$ $f_c(x) = \sqrt{100-x^2},~x\in [-10,10]$ $f^\prime_c(x) = -\dfrac{x}{\sqrt{100-x^2}}$ 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.
 October 30th, 2017, 10:30 PM #3 Newbie   Joined: Oct 2017 From: us Posts: 3 Thanks: 0 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
October 30th, 2017, 10:53 PM   #4
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 Originally Posted by ezmoney 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 10:56 PM.

October 30th, 2017, 11:15 PM   #5
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 Originally Posted by romsek 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 $\sqrt{100-x^2}$

October 30th, 2017, 11:51 PM   #6
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 Originally Posted by ezmoney 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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