A woman is attached to a bungee cord from a bridge that is 38m above a river. Her height in meters above the river t seconds after the jump is y(t)=19(1 + e^-1xcos(t)) for t>0.
What is her velocity at t=1 and t=4?
I am wondering what the x variable is? or is that times?
t=0 initially
I assume you meant
y(t) = 19(1+e^(-cos(t))
so,
v(t) = 19sin(t) e^(-cos(t))
v(1) = 19 sin(1) e^-cos(1) = 9.31
v(4) = 19 sin(4) e^-cos(4) = -27.64
To find the woman's velocity at t = 1 and t = 4, we need to take the derivative of the height function with respect to time.
The given height function is y(t) = 19(1 + e^(-1x)cos(t)).
To find the derivative, we need to use the chain rule since there is a composition of functions.
Let's break down the function first:
y(t) = 19(1 + e^(-x)cos(t))
Now, we can apply the chain rule:
dy/dt = d/dt [19(1 + e^(-x)cos(t))]
Using the chain rule, we differentiate each part separately:
d(1 + e^(-x)cos(t))/dt
Differentiating the constant 1 with respect to t gives us 0.
Taking the derivative of e^(-x)cos(t) with respect to t requires the product rule:
d(e^(-x))/dt * cos(t) + e^(-x) * d(cos(t))/dt
Now, let's differentiate each part:
d(e^(-x))/dt = -e^(-x) * dx/dt
d(cos(t))/dt = -sin(t)
Substituting these derivatives back into the equation:
dy/dt = 0 + (-e^(-x) * dx/dt) * cos(t) + e^(-x) * (-sin(t))
Now, we need to find dx/dt, the derivative of x with respect to t. However, the problem only provides the height function, y(t). To find dx/dt, we would need additional information.
Without the value or equation for x(t), we cannot calculate the derivative dx/dt, and therefore, we cannot determine the woman's velocity at t = 1 and t = 4.