$A = 5\sin120\pi t$, find rate of change at $t = 1s$.

Question

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.

Answer 1

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:$$f'(x)=\lim_{h\to 0} {f(x+h)-f(x)\over h}$$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$$f(x)=5\sin 120\pi t\to f'(x)=600\pi \cos 120\pi t\to f'(1)=600\pi\approx 1884.95559215$$

Answer 2

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

Answer 3

$A = 5 sin120\pi t$.

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

Evaluated at $t=1s$, the rate of change of current is $600\pi cos120\pi = 600 \pi \;A/s$.

Answer 4

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

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

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

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

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

$$\color{blue}{t = 1} \implies A’(1) = 600\pi\cos(120\pi\cdot\color{blue}{1}) = 600\pi\cos(120\pi)$$

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

$$A’(1) = 600\pi\cdot 1 = 600\pi$$