Htykuut

I'm in grade 12 advanced functions. Here's the question I'm having trouble with;

The current in a household appliance varies according to the equation $A = 5 sin 120 pi t$, where $A$ is the current, in amperes, and $t$ is the time, in seconds. At what rate is the current changing at $t = 1s$?

Any help? Thanks.

The definitions of variation and rate formall go back to the definition of 1st-order derivative. For any 1st-order differentiable function $f(x)$, the 1st-order derivation is defined as below:$$f'(x)=lim_{hto 0} {f(x+h)-f(x)over h}$$which is the slope of the tangent to the function in point $(x,f(x))$. It also shows how approximately does a function look like near a point.

In this case, the rate can be calculated as following$$f(x)=5sin 120pi tto f'(x)=600pi cos 120pi tto f'(1)=600piapprox 1884.95559215$$

The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.

$A = 5 sin120pi t$.

$frac{dA}{dt} = 600pi cos 120 pi t$

Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.

$$A(t) = 5sin (120pi t)$$

Differentiate $A(t)$ using the Chain Rule.

$$frac{dA}{dt} = 5cos(120pi t)cdot 120pi$$

$$frac{dA}{dt} = 600picos(120pi t)$$

So, the derivative found gives the instantaneous rate at $t$ seconds. In this case, let $t = 1$.

$$color{blue}{t = 1} implies A’(1) = 600picos(120picdotcolor{blue}{1}) = 600picos(120pi)$$

Recall $cos(pi n) = 1$ for all even integers $n$.

$$A’(1) = 600picdot 1 = 600pi$$