To find the rate of change of $T$ with respect to $t$, we need to compute the derivative of $T$ with respect to $t$, denoted as $\frac{dT}{dt}$.

Given: $T(t) = 10 \left(\frac{4t^2 + 16t + 85}{t^2 + 4t + 10}\right)$

Step 1: Differentiate $T(t)$ with respect to $t$

We can rewrite $T(t)$ as: $T(t) = 10 \cdot g(t)$ where $g(t) = \frac{4t^2 + 16t + 85}{t^2 + 4t + 10}$

Using the product rule and the chain rule: $\frac{dT}{dt} = 10 \cdot \frac{dg(t)}{dt}$

Now, we differentiate $g(t)$ using the quotient rule. The quotient rule states: $\frac{d}{dt} \left(\frac{u(t)}{v(t)}\right) = \frac{v(t) \cdot u'(t) - u(t) \cdot v'(t)}{[v(t)]^2}$ where:

• $u(t) = 4t^2 + 16t + 85$
• $v(t) = t^2 + 4t + 10$

First, compute the derivatives of $u(t)$ and $v(t)$: $u'(t) = \frac{d}{dt}(4t^2 + 16t + 85) = 8t + 16$ $v'(t) = \frac{d}{dt}(t^2 + 4t + 10) = 2t + 4$

Now, apply the quotient rule: $\frac{dg(t)}{dt} = \frac{(t^2 + 4t + 10)(8t + 16) - (4t^2 + 16t + 85)(2t + 4)}{(t^2 + 4t + 10)^2}$

Step 2: Simplify and evaluate at $t = 10$

Substitute $t = 10$ into the equation:

• $u(10) = 4(10)^2 + 16(10) + 85 = 400 + 160 + 85 = 645$
• $v(10) = (10)^2 + 4(10) + 10 = 100 + 40 + 10 = 150$
• $u'(10) = 8(10) + 16 = 80 + 16 = 96$
• $v'(10) = 2(10) + 4 = 20 + 4 = 24$

Substitute these values into the derivative: $\frac{dg(10)}{dt} = \frac{150 \cdot 96 - 645 \cdot 24}{150^2}$ Simplify this expression.

Finally, multiply by 10 to get $\frac{dT}{dt}$ at $t = 10$: $\frac{dT}{dt} = 10 \cdot \frac{dg(10)}{dt}$

Let's calculate this.The rate of change of $T$ with respect to $t$ when $t = 10$ is $-0.48$.

Would you like more details or have any questions about this process?

Here are 5 related questions:

1. How do you apply the product rule when differentiating functions?
2. What are the conditions for using the quotient rule in differentiation?
3. How can we interpret the sign of the rate of change in a physical context?
4. What are some alternative methods for finding the derivative of a rational function?
5. How would the result change if we considered a different value of $t$?

Tip: When dealing with complex derivatives, breaking down the function into smaller parts and applying differentiation rules step-by-step can simplify the process significantly.