If S=(2+3t)(5t-4),find ds/dt when t=4/5sec.
s'=(2+3t)(5)+ (5t-4)(3)
put in t=.8, and solve.
thpo;o
To find ds/dt, we need to differentiate the expression S with respect to t.
Given that S = (2+3t)(5t-4), we can use the product rule of differentiation to find ds/dt.
The product rule states that if we have two functions u(t) and v(t), then the derivative of their product is given by:
(d/dt)(u(t) * v(t)) = u(t) * (d/dt)v(t) + v(t) * (d/dt)u(t)
Let's apply the product rule to find ds/dt:
S = (2 + 3t)(5t - 4)
Taking the derivative of S with respect to t:
(d/dt)S = (2 + 3t) * (d/dt)(5t - 4) + (5t - 4) * (d/dt)(2 + 3t)
Now, let's find the derivatives:
(d/dt)(5t - 4) = 5 (since the derivative of 5t with respect to t is 5, and the derivative of -4 with respect to t is 0)
(d/dt)(2 + 3t) = 3 (since the derivative of 2 with respect to t is 0, and the derivative of 3t with respect to t is 3)
Substituting these results back into our previous equation:
(d/dt)S = (2 + 3t) * 5 + (5t - 4) * 3
Simplifying further:
(d/dt)S = 10 + 15t + 15t - 12
(d/dt)S = 30t - 2
Now that we have the derivative (ds/dt), we can substitute the value of t and calculate it when t = 4/5 sec:
(ds/dt) = 30(4/5) - 2
(ds/dt) = 24/sec - 2
(ds/dt) = 22/sec
Therefore, when t = 4/5 sec, ds/dt is equal to 22/sec.