Question

If \[2t = {v^2}\] then what is the value of \[\dfrac{{dv}}{{dt}}\]?
A. \[0\]
B. \[\dfrac{1}{4}\]
C. \[\dfrac{1}{2}\]
D. \[\dfrac{1}{v}\]

Answer 1

Hint: We have given the function and we have to find the derivative of the function. We will write v in terms of t and then will take the derivative with respect to t.

Formula used: \[\dfrac{{d\left( {{y^n}} \right)}}{{dx}} = n{y^{n - 1}}\dfrac{{dy}}{{dx}}\]

Complete step by step solution: Given: \[\] \[2t = {v^2}\]
\[ \Rightarrow v = \pm \sqrt {2t} \] (1.1)
Take derivative on the both sides in the equation (1.1) –
\[\dfrac{{dv}}{{dt}} = \pm \dfrac{{d\left( {\sqrt {2t} } \right)}}{{dt}}\]
Applying the formula of differentiation \[\dfrac{{d\left( {{y^n}} \right)}}{{dx}} = n{y^{n - 1}}\dfrac{{dy}}{{dx}}\] in above relation
\[\dfrac{{dv}}{{dt}} = \pm \sqrt 2 \left( {\dfrac{1}{2}} \right){t^{\dfrac{{ - 1}}{2}}}\dfrac{{dt}}{{dt}}\]
\[ \Rightarrow \dfrac{{dv}}{{dt}} = \pm \dfrac{1}{{\sqrt {2t} }}\]
But by equation (1.1) \[v = \pm \sqrt {2t} \]
 ∴ \[\dfrac{{dv}}{{dt}} = \pm \dfrac{1}{v}\]

So, Option ‘D’ is correct

Additional Information: The derivative of a curve represents the slope of the tangent of the curve. It is also referred to as the rate of change of the curve with respect to the variable.
The reverse process of the derivative is integration.

Note: If \[{x^2} = a \Rightarrow x = \pm \sqrt a \] (students makes mistakes there while taking only positive values which will be an incomplete answer ,for getting complete answer both values must be considered )hence there will be two values of the derivative. One with the positive sign and the other with the negative sign but since the option given is only contained with positive value so we will choose that option.