Let a smooth curve y=f(x) be such that the slope of the tangent at any point (x, y) on it …

Question

Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $$\left(\frac{-y}{x}\right)$$. If the curve passes through the points $(1,2)$ and $(8,1)$, then $$\left|y\left(\frac{1}{8}\right)\right|$$ is equal to

Accepted Answer

$${{dy} \over {dx}} \propto {{ - y} \over x}$$

$${{dy} \over {dx}} = {{ - ky} \over x} \Rightarrow \int {{{dy} \over y} =  - K\int {{{dx} \over x}} } $$

$$\ln |y| =  - K\ln |x| + C$$

If the above equation satisfy (1, 2) and (8, 1)

$$\ln 2 =  - K \times 0 + C \Rightarrow C = \ln 2$$

$$\ln 1 =  - K\ln 8 + \ln 2 \Rightarrow K = {1 \over 3}$$

So, at $$x = {1 \over 8}$$

$$\ln |y| =  - {1 \over 3}\ln \left( {{1 \over 8}} \right) + \ln 2 = 2\ln 2$$

$|y| = 4$

Solutions · 1