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Related rates differentiation problem

 
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Jul27-12, 11:32 PM   #18
 

Related rates differentiation problem


35 * 4 = 140 for x, 25 * 4 = 100 for y; [itex]\frac{-2(150-140)*35+2*100*25}{\sqrt{(150-140)^{2}+ 100^{2}}}[/itex] = [itex]\frac{-2*10*35 + 5000}{\sqrt{10^{2} + 100^{2}}}[/itex] = [itex]\frac{-700 + 5000}{\sqrt{10100}}[/itex] = [itex]\frac{4300}{10\sqrt{101}}[/itex]
Jul27-12, 11:35 PM   #19
 
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Quote by frosty8688 View Post
35 * 4 = 140 for x, 25 * 4 = 100 for y; [itex]\frac{-2(150-140)*35+2*100*25}{\sqrt{(150-140)^{2}+ 100^{2}}}[/itex] = [itex]\frac{-2*10*35 + 5000}{\sqrt{10^{2} + 100^{2}}}[/itex] = [itex]\frac{-700 + 5000}{\sqrt{10100}}[/itex] = [itex]\frac{4300}{10\sqrt{101}}[/itex]
Looks like you removed the 2 in the denominator, but never factored out or canceled the 2's in the numerator. Let's just forget about canceling out the 2. You should have written:
[tex]\frac{-2(150-140) \cdot 35+2 \cdot 100 \cdot 25}{2\sqrt{(150-140)^{2}+ 100^{2}}}[/tex]
(Also, don't use "*" for multiplication in LaTeX. Use "\cdot" instead.)
Jul27-12, 11:37 PM   #20
 
I see, my mistake. I must have missed the 2 in the denominator.
Jul27-12, 11:39 PM   #21
 
It would be about 21.39 km/hr
Jul27-12, 11:43 PM   #22
 
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That's what I got. You have to be careful with the algebra when solving Calculus problems. (Not easy -- I know.)
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