2-dimensional bowl

Calculus Level 2

The ends $A$ and $B$ of a rod of length of $\sqrt{5}$ are sliding along the curve $y=2x^2$ . Let $x_A$ and $x_B$ be the $x$ -coordinates of the ends. At the moment when $A$ is at $(0,0)$ and $B$ is at $(1,2)$ . Find the value of the derivative $\dfrac{dx_B}{dx_A}$ .

\frac{1}{3}

\frac{1}{\sqrt{5}}

9

\frac{1}{9}

1 solution

Tom Engelsman
Nov 2, 2016

Let a = x a, b = x b, and d(x b)/d(x a) = db/da = b'. The length of the rod is expressible as:

5 = (b - a)^2 + (2 b^2 - 2 a^2)^2 = (b - a)^2 + 4*(b^2 - a^2)^2 (i)

and implicitly differentiating both sides of (i) with respect to a yields:

0 = 2 (b-a) (b' - 1) + 8 (b^2 - a^2) (2bb' - 2a);

or 2 (b-a) + 16a (b^2 - a^2) = [2 (b-a) + 16b (b^2 - a^2)]*b';

or b' = [2 (b-a) + 16a (b^2 - a^2)] / [2 (b-a) + 16b (b^2 - a^2)] (ii)

A final substitution of a = x a = 0 and b = x b = 1 into (ii) gives b' = 2/(2 +16) = 1/9, or choice D as a correct answer.

Thanks for adding a solution to this old problem!

Calvin Lin Staff - 4 years, 7 months ago

My pleasure, Calvin! I absolutely love Brilliant, and I do enjoy providing solutions where needed :)

tom engelsman - 4 years, 7 months ago

Great! If you would like more problems, here is a list of problems that we're looking for solutions, so that we can feature them. It would be great if you can chip in for any area that you're comfortable in.

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Calvin Lin Staff - 4 years, 7 months ago

2-dimensional bowl

1 solution

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