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), dx/dt=6. What is the value of dy/dt at this time?

Answer 1

Answer:

$\displaystyle y'=-\frac{9}{2}$

Step-by-step explanation:

Differentiation

We have a relationship between x and y as follows:

$x^2+y^2=25$

Both variables depend on time t for t≥0.

Differentiating with respect to time:

$(x^2)'+(y^2)'=(25)'$

Applying the derivative of a power function:

$2xx'+2yy'=0$

Recall the derivative of a constant is 0.

Dividing by 2:

$xx'+yy'=0$

Solving for y':

$\displaystyle y'=-\frac{xx'}{y}$

At some specific time, we have:

x=3, y=4, dx/dt = x' = 6. Substituting:

$\displaystyle y'=-\frac{3\cdot 6}{4}$

Operating:

$\displaystyle y'=-\frac{18}{4}$

Simplifying:

$\mathbf{\displaystyle y'=-\frac{9}{2}}$