For the curve y=5x-2x^{3},if x increases at t | vedclass

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

Question

(A) The slope of the curve is given by $m = \frac{dy}{dx} = 5 - 6x^{2}$.To find the rate of change of the slope with respect to time $t$,we differenti

Vedclass · Accepted Answer

-$72$ units/sec

Vedclass · Answer

(A) The slope of the curve is given by $m = \frac{dy}{dx} = 5 - 6x^{2}$.To find the rate of change of the slope with respect to time $t$,we differentiate $m$ with respect to $t$ using the chain rule:$\frac{dm}{dt} = \frac{d}{dt}(5 - 6x^{2}) = -12x \cdot \frac{dx}{dt}$.Given that $x = 3$ and $\frac{dx}{dt} = 2 	ext{ units/sec}$,we substitute these values into the equation:$\frac{dm}{dt} = -12(3)(2) = -72 	ext{ units/sec}$.Thus,the slope of the curve is changing at the rate of $-72 	ext{ units/sec}$,which means it is decreasing at a rate of $72 	ext{ units/sec}$.

For the curve $y=5x-2x^{3}$,if $x$ increases at the rate of $2 \text{ units/sec}$,then how fast is the slope of the curve changing when $x=3$?

Similar Questions

$A$ man $2 \, m$ high walks at a uniform speed of $5 \, m/h$ away from a lamp post $6 \, m$ high. The rate at which the length of his shadow increases is

$A$ spherical balloon is expanding. If at any instant the rate of increase of its volume is $16$ times the rate of increase of its radius,then its radius at that instant 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 $km/h$):

The volume of a ball is increasing at the rate of $4 \pi \text{ cc/sec}$. The rate of increase of the radius,when the volume is $288 \pi \text{ cc}$,is

The radius of a circle is increasing uniformly at the rate of $3 \, cm/s$. The rate of increase of its area when the radius is $10 \, cm$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module