Question

How do you find the value of $t$ for which $\dfrac{{dy}}{{dx}} = 0$ , given $x = 80t$ and $y = 64t - 16{t^2}$ ?

Answer 1

In this question, it is observable that differentiation is being originally performed with a different variable than that in which the expressions are given. Actually, the expression variable is used, but it is not visible, because both x and y were differentiated with respect to t. Students need to first find the derivatives of x and y with respect to t, to proceed further.

Complete step by step answer:
We have,

$$\dfrac{{dy}}{{dx}} = 0 \\\ \Rightarrow x = 80t \\\ \Rightarrow y = 64t - 16{t^2} \\\$$

We need to find the value of $t$ for which $\dfrac{{dy}}{{dx}} = 0$ .
To solve this question, we need to figure out the differentiation of $x$ and $y$ with respect to $t$ , and then use that to calculate the value of $t$ .
Now,
$\dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {80t} \right)$
On differentiating, we get $\dfrac{{dx}}{{dt}} = 80$
Similarly, we will differentiate $y$ .
We have,
$\dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left( {64t - 16{t^2}} \right)$
On differentiating, we get $\dfrac{{dy}}{{dt}} = 64 - 32t$
Now, $\dfrac{{dy}}{{dx}}$ can also be written as
$\dfrac{{dy}}{{dx}} = \dfrac{{dy/dt}}{{dx/dt}} = 0$
Therefore, on putting the values calculated above, we get

$$\dfrac{{dy}}{{dx}} = \dfrac{{64 - 32t}}{{80}} = 0 \\\ \Rightarrow64 - 32t = 0 \\\$$

Thus, on solving the obtained equation, we will get $t = 2$

Hence, the value of $t$ for which $\dfrac{{dy}}{{dx}} = 0$ , given $x = 80t$ and $y = 64t - 16{t^2}$ is 2

Note: In the field of Mathematics, Differentiation can be defined as a derivative of a function with respect to an independent variable. Differentiation in calculus, can be applied to measure the function per unit change in any independent variable. The best examples of Differentiation in real life are calculations of instantaneous velocity and the rate of change of any physical quantity with respect to time.
There are various identities that can be applied to make differentiation problems easier to solve. Students are advised to learn how to apply them.