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:

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

Answer 2

The movement of the particle on the circle is its displacement

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

The equation of the circle is given as:

$\mathbf{x^2 + y^2 = 25}$

Differentiate with respect to time

$\mathbf{2x\times \frac{dx}{dt} + 2y \times \frac{dy}{dt} = 0}$

Divide through by 2

$\mathbf{x\times \frac{dx}{dt} + y \times \frac{dy}{dt} = 0}$

At point (3,4);

x =3 and y = 4.

So, we have:

$\mathbf{3 \times \frac{dx}{dt} + 4 \times \frac{dy}{dt} = 0}$

Also: dx/dt = 6

So, we have:

$\mathbf{3 \times 6 + 4 \times \frac{dy}{dt} = 0}$

Multiply

$\mathbf{18+ 4 \times \frac{dy}{dt} = 0}$

Subtract 18 from both sides

$\mathbf{4 \times \frac{dy}{dt} = -18}$

Divide through by 4

$\mathbf{\frac{dy}{dt} = -\frac{18}4}$

Simplify

$\mathbf{\frac{dy}{dt} = -\frac{9}2}$

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

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