Question

A curve has the equation y = x/16 * (5-x)^4. Calculates the values of x for which dy/dx =0 . Given that a small change,p,is made in the x-coordinate at the point (4,0.25), calculate,in terms of p, the approximate change in the y-coordinate .

Answer 1

Answer on Question #42962, Math, Differential Geometry

A curve has the equation $y = x / 16 * (5 - x)^4$. Calculates the values of $x$ for which $\frac{dy}{dx} = 0$. Given that a small change, $p$, is made in the $x$-coordinate at the point $(4,0.25)$, calculate, in terms of $p$, the approximate change in the $y$-coordinate.

Answer.

$$y = \frac{x (5 - x)^4}{16} \rightarrow \frac{dy}{dx} = \frac{(5 - x)^4}{16} - \frac{4x (5 - x)^3}{16} = \frac{5(1 - x)(5 - x)^3}{16}$$

So, $\frac{dy}{dx} = 0$ when $x = 1$ or $x = 5$.

Let $x = 4 + p \rightarrow y(4 + p) = \frac{(4 + p)(1 - p)^4}{16}$.

For small $p$, $y(4 + p) = \frac{4 - 15p}{16} + O(p^2) = 0.25 - \frac{15}{16} p + O(p^2)$.

Therefore, the approximate change in the $y$-coordinate is: $\Delta y = -\frac{15}{16} p$.

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