Chapter 4, Problem 60RE

To determine

To find : The value of $dx/dt$ when $x=3$ .

Answer to Problem 60RE

The answer:

  $\frac{dx}{dt}=4\text{ units/sec }$

Explanation of Solution

Given information:

The curve $y=x^{3/2}$ in the first quadrant.

The origin increase at the constant rate of $11$ units per second.

Calculation:

Particle at point $y=x^{3/2}$, moving along the curve:

  

At the moment,

  $\begin{array}{c}\text{x=3}\\ y=3^{3/2}\\ =\sqrt{27}\\ h=\sqrt{x^2+y^2}\\ =\sqrt{9+27}\\ =\sqrt{36}\\ =6\end{array}$

Implicitly differentiate with respect to $t$ , plug in all the known values, then , solve for $\frac{dx}{dt}$:

  $\begin{array}{c}\frac{dh}{dt}=11\text{ units }/\text{sec}\\ y=x^{3/2}\\ h^2=x^2+y^2\\ =x^2+{\left(x^{3/2}\right)}^2\\ =x^2+x^3\\ 2h\frac{dh}{dt}=\left(2x+3x^2\right)\frac{dx}{dt}\\ \Rightarrow 2(6)(11)=\left(2(3)+3{(3)}^2\right)\frac{dx}{dt}\\ \Rightarrow \frac{dx}{dt}=\frac{132}{33}\\ =4units/\sec \end{array}$

Therefore, the required value of $dx/dt$ when $x=3$ is

  $\frac{dx}{dt}=4\text{ units/sec }$ .