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 February 21st, 2011, 07:01 PM #1 Senior Member   Joined: Feb 2011 Posts: 118 Thanks: 0 Derivatives: Differential and Linear approximation HELP Hi, I got a review sheet, and I'm having trouble solving these particular questions on differentials and linear approximations...I need some assistance please. 1) In a manufacturing process, ball bearings must be made with radius of 0.5 mm, with a maximum error in the radius of (+or-)0.018 mm. Estimate the maximum error in the volume of the ball bearing. The formula for the volume of the sphere is_______. If an error (delta)r is made in measuring the radius of the sphere, the maximum error in the volume is (delta)V=______________. Rather than calculating (delta)V, approximate (delta)V with dV, wher dV=_________. Replacing r with _________ and dr=(delta)r with (+ or -)_________ gives dV= (+ or -)_________ The maximum error in the volume is about____________mm(cubed). 2)The demand function for a product is given by p=f(q)=90?sqrt(q). Where p is the price per unit in dollars for q units. Use the linear approximation to approximate the price when 2024 units are demanded. Solution: We want to approximate f(2024). From: f(q)*approximate symbol*=L(q)=f(a)+f'(a)(q-a) and the fact that f'(a)___________(-1/sqrt(a) or 1/2(sqrt(a)) or -1/2(sqrt(a)) or 1/sqrt(a) ) we choose a=_________. From f(2025)=_______ and f'(2025)=__________ we get f(2024)*approximate symbol*_________. hence, the price per unit when 2024 units are demanded is approximately \$________. IF anything is unclear on my part, describing the problems, please let me know.. Thank you.
 February 22nd, 2011, 08:07 PM #2 Senior Member   Joined: Feb 2011 Posts: 118 Thanks: 0 Re: Derivatives: Differential and Linear approximation HELP I have figured out the second question that is given. So I'm only still stuck on the first question. Thanks!
 February 22nd, 2011, 09:29 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Derivatives: Differential and Linear approximation HELP The formula for the volume V of a sphere of radius r is: $V=\frac{4}{3}\pi r^3$ Thus, $\Delta V=V$$r+\Delta r$$-V(r)=\frac{4}{3}\pi$$\(r+\Delta r$$^3-r^3\)=$ $\frac{4}{3}\pi$$3r^2\Delta r+3r\(\Delta r$$^2+$$\Delta r$$^3\)$ and $dV=4\pi r^2dr$ Replacing r with 0.5 mm and dr = ?r with ±0.018 mm gives: $dV=4\pi $$0.5\text{ mm}$$^2$$\pm0.018\text{ mm}$$=\pm0.018\pi\text{ mm^3}$ Thus, the maximum error is about 0.0565 mm³. Thanks from Yuliecw
 February 23rd, 2011, 04:41 PM #4 Senior Member   Joined: Feb 2011 Posts: 118 Thanks: 0 Re: Derivatives: Differential and Linear approximation HELP Thanks a lot!
February 24th, 2011, 06:08 PM   #5
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Re: Derivatives: Differential and Linear approximation HELP

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 Originally Posted by ProJO I have figured out the second question that is given. So I'm only still stuck on the first question. Thanks!
How did you do the 2nd one?? thanks

 February 24th, 2011, 06:28 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Derivatives: Differential and Linear approximation HELP We may use the linear approximation to f(q) by the point-slope formula: $f(q)\approx L(q)=f#39;(a)(q-a)+f(a)$ With $f(q)=90-\sqrt{q}$ we have $f'(q)=-\frac{1}{2}q^{-\frac{1}{2}}=-\frac{1}{2\sqrt{q}}$ so $f'(a)=-\frac{1}{2\sqrt{a}}$ and since $45^2=2025$ and $2024\approx2025$ we choose a = 2025. Thus with: $f(2025)=90-\sqrt{2025}=90-45=45$ and $f'(2025)=-\frac{1}{2\sqrt{2025}}=-\frac{1}{2\cdot45}=-\frac{1}{90}$ we have: $f(2024)\approx-\frac{1}{90}(2024-2025)+45=\frac{4051}{90}\approx45.01$ This is very close to the actual value of f(2024).

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# where p is the price per unit in dollars for q units. Use the linear approximation to approximate the price when 2499 units are demanded

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