Question

A particle moves on the circle x2 y2=25 in the xy-plane for time t≥0. At the time when the particle is at the point (3,4), dxdt=6. What is the value of dydt at this time?

Answer 1

The movement of the particle on the circle is its displacement.

The value of dy/dt at this time is -9/2.

What is the differentiation?

Differentiation is a process, in Maths, where we find the instantaneous rate of change in function based on one of its variables.

A particle moves on the circle x^2 + y^2=25 in the XY-plane for time t≥0. At the time when the particle is at the point (3,4), dxdt=6.

The equation of the circle is given as:

$\rm x^2 + y^2=25$

Differentiate with respect to time

$\rm 2x \times \dfrac{dx}{dt}+2y \times \dfrac{dy}{dt}=0$

Substitute all the values in the equation

$\rm 2x \times \dfrac{dx}{dt}+2y \times \dfrac{dy}{dt}=0\\\\2(3) \times 6+2(4)\times \dfrac{dy}{dt}=0\\\\ 6 \times 6+8 \times \dfrac{dy}{dt}=0\\\\ 36+8\times \dfrac{dy}{dt}=0\\\\ 8\times \dfrac{dy}{dt}=-36\\\\ \dfrac{dy}{dt}= \dfrac{-36}{8}\\\\ \dfrac{dy}{dt}= \dfrac{-9}{2}$

Hence, the value of dy/dt at this time is -9/2.

Learn more about differentiation here;

brainly.com/question/19385433

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