Question

Functions w, x, and y are differentiable with respect to time and are related by the equation w = rx. If x is decreasing at a constant rate of 1 unit per minute and y is increasing at a constant rate of 4 units per minute, at what rate is w changing with respect to time when x = 6 and y = 20? a) -384 b) -240 c) 96 d) 276

Answer 1

To find the rate at which w is changing concerning time, we need to differentiate the equation w = rx. Using the product rule and the chain rule, we can find the derivative of w concerning time and substitute the given values to solve for the rate of change. The answer is -384.

Step-by-step explanation:

We are given the equation w = rx, where w, x, and y are differentiable functions concerning time. We are also given that x is decreasing at a constant rate of 1 unit per minute and y is increasing at a constant rate of 4 units per minute.

To find the rate at which w is changing concerning time, we need to differentiate the equation w = rx concerning time. Since w and x are both functions of time, we can use the chain rule to differentiate the equation.

d/dt(w) = d/dt(rx)

Using the product rule and the chain rule, we have:

d/dt(w) = x * d/dt(r) + r * d/dt(x)

Since x is decreasing at a constant rate of 1 unit per minute, we have d/dt(x) = -1. Given that y is increasing at a constant rate of 4 units per minute, this means that d/dt(y) = 4. Since w = rx, we can substitute the given values to find the rate at which w is changing concerning time when x = 6 and y = 20.

d/dt(w) = 6 * d/dt(r) + r * (-1)

Substituting the values x = 6 and y = 20, we can solve for d/dt(w).

Answer 2

The rate at which w is changing with respect to time when x = 6 and y = 20 is 276 units per minute.

Step-by-step explanation:

We are given that w = rx, where r is a constant. This means that w is directly proportional to x. Since x is decreasing at a constant rate of 1 unit per minute, we can say that r is negative.

Let's find the derivative of w with respect to time:

dw/dt = r(dx/dt)

Substituting the values given, we get:

dw/dt = -r(1) = -r

Since y is increasing at a constant rate of 4 units per minute, we can say that y = kx, where k is a constant. This means that y is directly proportional to x. Substituting the values given, we get:

20 = k(6)

k = 3.33333333333333 (rounded to 10 decimal places)

Now, let's find the derivative of y with respect to time:

dy/dt = k(dx/dt)

Substituting the values given, we get:

dy/dt = 3.3333333333333(1) = 3.33 units per minute (rounded to nearest hundredth)

We know that w = rx and x is decreasing at a constant rate of 1 unit per minute. So, dw/dt = -r. When x = 6 and y = 20, we can find the value of r as follows:

20 = -r(6)

r = -276 (rounded to nearest hundredth)

Substituting this value of r in dw/dt, we get:

dw/dt = -(-276) = 276 units per minute (rounded to nearest hundredth)