A point on the parabola y^2 = frac{36}{5}x at | vedclass

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

Question

(B) Given the parabola equation: $y^2 = \frac{36}{5}x$. Differentiating both sides with respect to time $t$: $2y \frac{dy}{dt} = \frac{36}{5} \frac{dx

Vedclass · Accepted Answer

$(1/5, 6/5)$

Vedclass · Answer

(B) Given the parabola equation: $y^2 = \frac{36}{5}x$. Differentiating both sides with respect to time $t$: $2y \frac{dy}{dt} = \frac{36}{5} \frac{dx}{dt}$. We are given that the ordinate $(y)$ increases at thrice the rate of the abscissa $(x)$, so $\frac{dy}{dt} = 3 \frac{dx}{dt}$. Substituting this into the differentiated equation: $2y(3 \frac{dx}{dt}) = \frac{36}{5} \frac{dx}{dt}$. Assuming $\frac{dx}{dt} 
eq 0$, we get $6y = \frac{36}{5}$, which simplifies to $y = \frac{6}{5}$. Now, substitute $y = \frac{6}{5}$ into the original parabola equation: $(\frac{6}{5})^2 = \frac{36}{5}x$. $\frac{36}{25} = \frac{36}{5}x$. Solving for $x$: $x = \frac{36}{25} 	imes \frac{5}{36} = \frac{1}{5}$. Thus, the required point is $(1/5, 6/5)$.

$A$ point on the parabola $y^2 = \frac{36}{5}x$ at which the ordinate increases at thrice the rate of the abscissa is:

Similar Questions

If the sides of a triangle vary slightly in such a way that its circumradius remains constant,then $\frac{da}{\cos A} + \frac{db}{\cos B} + \frac{dc}{\cos C}$ is equal to

Gas is being pumped into a spherical balloon at the rate of $30 \ ft^3 / \text{min}$. The rate at which the radius increases when it reaches the value $15 \ ft$,is:

$A$ balloon which always remains spherical is being inflated by pumping in $10 \text{ cm}^3$ of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is $15 \text{ cm}$.

$A$ ladder is resting against a wall at an angle of $30^\circ$ with the wall. $A$ man is ascending the ladder at the rate of $3 \, ft/sec$. His rate of approaching the wall is:

$A$ spherical iron ball of radius $10 \, cm$ is coated with a layer of ice of uniform thickness that melts at a rate of $50 \, cm^3/min$. When the thickness of the ice is $5 \, cm$,find the rate at which the thickness of the ice decreases.

Vedclass Test Series

Exam Paper Generator

Online Exam Module