Question:

At ........... point on the parabola $x^2 = 4y$. the rate of increase of the x-coordinate is the same as the rate of the increasing y-coordinate.

Updated On: Jul 6, 2022
  • $(2, 1)$
  • $(- 3, 1)$
  • $(- 2, 1/4)$
  • $(7/4, 1/4)$
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The Correct Option is A

Solution and Explanation

As given $\frac{dx}{dt}=\frac{dy}{dt}$ Now $x^{2}=4y$ $\Rightarrow 2x \frac{dx}{dt}=4 \frac{dy}{dt}$ $\Rightarrow 2x=4$ $\Rightarrow x=2$ $\therefore 4y=\left(2\right)^{2}=4$ $\Rightarrow y=1$ $\therefore$ reqd. point is $\left(2, 1\right)$
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Notes on Application of derivatives

Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives