If the rate of change of x is more than the r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the curve $x^3 = 12y$. Differentiating both sides with respect to time $t$,we get: $3x^2 \frac{dx}{dt} = 12 \frac{dy}{dt}$ $\frac{dy}{dt} =

Vedclass · Accepted Answer

$(0, 2)$

Vedclass · Answer

(D) Given the curve $x^3 = 12y$. Differentiating both sides with respect to time $t$,we get: $3x^2 \frac{dx}{dt} = 12 \frac{dy}{dt}$ $\frac{dy}{dt} = \frac{3x^2}{12} \frac{dx}{dt} = \frac{x^2}{4} \frac{dx}{dt}$ Given that the rate of change of $x$ is more than the rate of change of $y$,i.e.,$\frac{dx}{dt} > \frac{dy}{dt}$. Substituting the value of $\frac{dy}{dt}$: $\frac{dx}{dt} > \frac{x^2}{4} \frac{dx}{dt}$ Since $x > 0$,$\frac{dx}{dt}$ is positive (assuming $x$ is increasing),so we can divide by $\frac{dx}{dt}$: $1 > \frac{x^2}{4}$ $x^2 $|x| Since the condition $x > 0$ is given,the interval for $x$ is $(0, 2)$.

If the rate of change of $x$ is more than the rate of change of $y$ on the curve $x^3 = 12y$ for $x > 0$,then $x$ lies in the interval:

Similar Questions

If the error committed in measuring the radius of the circle is $0.05 \%$,then the corresponding error in calculating the area is (in $\%$)

$A$ body moves according to the formula $v = 1 + t^2$,where $v$ is the velocity at time $t$. The acceleration after $3 \text{ s}$ will be .......... $\text{cm/s}^2$. ($v$ is in $\text{cm/s}$)

$A$ container is in the shape of an inverted cone. Its height is $6 \ m$ and the radius at the top is $4 \ m$. If it is filled with water at the rate of $3 \ m^3/min$,then the rate of change of the height of the water (in $m/min$) when the water level is $3 \ m$,is

The volume of a spherical balloon is increasing at the rate of $2 \ cm^3/sec$. When its radius is $4 \ cm$, the rate of change of its surface area (in $cm^2/sec$) is

If a man of height $1.8 \ m$ is walking away from the foot of a light pole of height $6 \ m$ with a speed of $7 \ km/h$ on a straight horizontal road,then the rate of change of the length of his shadow is (in $km/h$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module