4

When $x=3$, the rate at which $\frac{1}{x}$ is decreasing is $k$ times the rate at which $x$ is increasing. What is the value of $k$?

To find the rate at which $\frac{1}{x}$ is decreasing, find the first derivative $\frac{-1}{x^2}$ Now this is to be $k$ the rate at which $x$ increases? Doesn't $x$ increase as a linear function, i.e. the coefficient which is one.

Thus at $x = 3$

$\frac{-1}{3^2} = k = -1/9$

I am asking because my solution is different from the answer sheet. The answer sheet involes $\frac{\mathrm{d}}{\mathrm{d}t}$, and I do not understand why.

Given that $\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{x}\right) = k\frac{\mathrm{d}x}{\mathrm{d}t}$, differentiate to get $-x^2\frac{\mathrm{d}x}{\mathrm{d}t} = k\frac{\mathrm{d}x}{\mathrm{d}t}$ and substitute $x=3$, to get $\frac{-1}{9}\frac{\mathrm{d}x}{\mathrm{d}t}=k\frac{\mathrm{d}x}{\mathrm{d}t}$. Solve to find $k=\frac{-1}{9}$

yiyi
  • 7,352

2 Answers2

5

The answer sheet uses $t$ for time. Then, rate at which $x$ changes with respect to time is $\frac{dx}{dt}$, and rate at which $\frac{1}{x}$ changes with respect to time is $\frac{d}{dt}\frac{1}{x}$.

If the rate at which $\frac{1}{x}$ decreases is $k$ times the rate at which $x$ increases, then $-\frac{d}{dt}\frac{1}{x}=k\frac{dx}{dt}$. Hence $k=\frac{1}{9}$.

1

Call a function $f(x)$, it has been defined such that the rate of $f(1/3)$ is decreasing at $k$ times the rate at which $f(3)$ is increasing. And rate of change is Derivative, in a mathematical function.It can generally be expressed as a ratio between a change in one variable relative to a corresponding change in another. That's why they have taken $\dfrac{d}{dt}$. (Usually, rates are taken with respect to time)

Inceptio
  • 7,881