Question

If $y=6x-{{x}^{3}}$ and x increases at the rate of 5 units per second, the rate of change of slope when x = 3 is:
(a) -90 units/sec
(b) 90 units/sec
(c) 180 units/sec
(d) -180 units/sec

Answer 1

Hint:First we will differentiate the given expression with respect to x and that will be the value of slope in terms of x now we will again differentiate it with respect to ‘t’ to find the rate of change of slope and we know the value of $\dfrac{dx}{dt}=5$ , using this we will find the value of $\dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)$ which will be the final answer.

Complete step-by-step answer:
Let’s start solving this question,
Now we will differentiate $y=6x-{{x}^{3}}$ with respect to x.
We know $\dfrac{d{(x^3)}}{dx}=3{{x}^{2}}$ and $\dfrac{d{(6x)}}{dx}=6$
So we get,
$\dfrac{dy}{dx}=6-3{{x}^{2}}$
Now we have found the value of slope in terms of x.
Now we will again differentiate it with respect to time to find the rate of change of slope.
$\dfrac{d{(-3x^2)}}{dx}=-3\dfrac{d{(x^2)}}{dx}=-3(2x)=-6x$
After differentiating with respect to t we get,
$\dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)=-6x\dfrac{dx}{dt}$
Now we know that $\dfrac{dx}{dt}=5$ and we need to find the value when x = 3,
Now substituting the values in $\dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)=-6x\dfrac{dx}{dt}$ we get,
$\dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)=-6\times 3\times 5=-90$
Hence, the rate of change of slope when x = 3 is -90 units/sec.
Hence option (a) is correct.

Note: Here we have differentiated the given expression two times and then substituted the value to find the answer, this step is very important. And one should know that the slope of any equation is $\dfrac{dy}{dx}$ , and that is what we have used here to solve this question.Students should remember the general differentiation formula of polynomial function i.e. $\dfrac{d{(x^n)}}{dx}=nx^{n-1}$ to solve these types of questions.