Rate Change Area Df7144

1. **State the problem:** We have a rectangle with length $L = 4 + t$ and height $H = t^6$. We want to find the rate of change of the area $A$ with respect to time $t$, i.e., $\frac{dA}{dt}$. 2. **Formula for area:** The area of a rectangle is given by: $$A = L \times H$$ 3. **Express area in terms of $t$:** $$A = (4 + t)(t^6) = t^6(4 + t)$$ 4. **Differentiate $A$ with respect to $t$ using the product rule:** Recall the product rule: $$\frac{d}{dt}[f(t)g(t)] = f'(t)g(t) + f(t)g'(t)$$ Let $f(t) = t^6$ and $g(t) = 4 + t$. Calculate derivatives: $$f'(t) = 6t^5$$ $$g'(t) = 1$$ Apply product rule: $$\frac{dA}{dt} = 6t^5(4 + t) + t^6(1) = 6t^5(4 + t) + t^6$$ 5. **Simplify the expression:** Expand the first term: $$6t^5 \times 4 = 24t^5$$ $$6t^5 \times t = 6t^6$$ So, $$\frac{dA}{dt} = 24t^5 + 6t^6 + t^6 = 24t^5 + 7t^6$$ 6. **Factor common terms:** $$\frac{dA}{dt} = t^5(24 + 7t)$$ 7. **Final answer:** The rate of change of the area with respect to time is: $$\boxed{\frac{dA}{dt} = t^5(24 + 7t)}$$ 8. **Compare with options:** None of the options exactly match $t^5(24 + 7t)$. However, the closest is Option D: $t^5(24 + 28t)$. Since the problem likely expects the derivative as $t^5(24 + 7t)$, the correct derivative is as above. **Note:** The options seem to have errors in coefficients. The correct derivative is $t^5(24 + 7t)$.