The Temperature T of food put in a freezer is
T = (700) / t^2+4t+10
where t is the time in hours. Find the rate of change of T with respect to t at each of the following times.
a. t=1
b. t=3
c. t=5
d. t=10
** i'll be able to do the others on my own if you please explain step by step how to do at least one of these. I have now idea where to start. SOMEONE PLEASE HELP ME A.S.A.P ! :(
THANK YOU.
you gotto differentiate the function
so if
T= 700* (t^-2) + 4t + 10
dT/dt = 700* (-2) (t^ -3) +4
now just plug in t = 1 in the above
same with the others too
To find the rate of change of T with respect to t, we need to find the derivative of the given temperature function, T = (700) / (t^2+4t+10), with respect to t. The derivative will give us the rate at which the temperature is changing.
To find the derivative, we can use the quotient rule. The quotient rule states that for a function of the form f(t) = g(t) / h(t), where g(t) and h(t) are differentiable functions, the derivative is given by:
f'(t) = (g'(t) * h(t) - g(t) * h'(t)) / (h(t))^2
Now, let's find the derivative of T = (700) / (t^2+4t+10) using the quotient rule.
Step 1: Find g(t) and g'(t)
g(t) = 700
g'(t) = 0 (since the derivative of a constant is 0)
Step 2: Find h(t) and h'(t)
h(t) = t^2 + 4t + 10
h'(t) = 2t + 4 (using the power rule and the derivative of a constant)
Step 3: Apply the quotient rule
T'(t) = (g'(t) * h(t) - g(t) * h'(t)) / (h(t))^2
T'(t) = (0 * (t^2 + 4t + 10) - 700 * (2t + 4)) / (t^2 + 4t + 10)^2
T'(t) = (-1400t - 2800) / (t^2 + 4t + 10)^2
Now, we have the derivative T'(t) of the temperature function T with respect to t.
To find the rate of change of T at specific times, substitute the given values of t into the derivative equation T'(t).
a. t = 1:
T'(1) = (-1400(1) - 2800) / ((1)^2 + 4(1) + 10)^2
T'(1) = (-1400 - 2800) / (1 + 4 + 10)^2
T'(1) = -4200 / (15)^2
T'(1) = -4200 / 225
T'(1) = -18.67 (approx)
b. t = 3:
T'(3) = (-1400(3) - 2800) / ((3)^2 + 4(3) + 10)^2
T'(3) = (-4200 - 2800) / (9 + 12 + 10)^2
T'(3) = -7000 / (31)^2
T'(3) = -7000 / 961
T'(3) = -7.28 (approx)
c. t = 5:
T'(5) = (-1400(5) - 2800) / ((5)^2 + 4(5) + 10)^2
T'(5) = (-7000 - 2800) / (25 + 20 + 10)^2
T'(5) = -9800 / (55)^2
T'(5) = -9800 / 3025
T'(5) = -3.24 (approx)
d. t = 10:
T'(10) = (-1400(10) - 2800) / ((10)^2 + 4(10) + 10)^2
T'(10) = (-14000 - 2800) / (100 + 40 + 10)^2
T'(10) = -16800 / (150)^2
T'(10) = -16800 / 22500
T'(10) = -0.75 (approx)
So, the rate of change of T with respect to t at each of the given times is approximately:
a. t = 1: -18.67
b. t = 3: -7.28
c. t = 5: -3.24
d. t = 10: -0.75