$A = 5sin120pi t$, find rate of change at $t = 1s$.
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.
calculus derivatives physics
edited Nov 25 at 7:58
jayant98
45715
asked Nov 25 at 6:06
Korvexius
42
add a comment |
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.
calculus derivatives physics
edited Nov 25 at 7:58
jayant98
45715
asked Nov 25 at 6:06
Korvexius
42
add a comment |
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.
calculus derivatives physics
edited Nov 25 at 7:58
jayant98
45715
asked Nov 25 at 6:06
Korvexius
42
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.
calculus derivatives physics
calculus derivatives physics
edited Nov 25 at 7:58
jayant98
45715
asked Nov 25 at 6:06
Korvexius
42
edited Nov 25 at 7:58
jayant98
45715
asked Nov 25 at 6:06
Korvexius
42
edited Nov 25 at 7:58
jayant98
45715
edited Nov 25 at 7:58
jayant98
45715
edited Nov 25 at 7:58
jayant98
45715
45715
asked Nov 25 at 6:06
Korvexius
42
asked Nov 25 at 6:06
Korvexius
42
asked Nov 25 at 6:06
Korvexius
42
42
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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$$f(x)=5sin 120pi tto f'(x)=600pi cos 120pi tto f'(1)=600piapprox 1884.95559215$$
answered Nov 25 at 9:01
Mostafa Ayaz
13.7k3836
add a comment |
The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.
answered Nov 25 at 6:46
Andrei
10.8k21025
add a comment |
$A = 5 sin120pi t$.
$frac{dA}{dt} = 600pi cos 120 pi t$
Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.
answered Nov 25 at 7:15
Aditya Dua
7868
add a comment |
$$A(t) = 5sin (120pi t)$$
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 $n$.
$$A’(1) = 600picdot 1 = 600pi$$
answered Nov 25 at 7:57
KM101
3,958417
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
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$$f(x)=5sin 120pi tto f'(x)=600pi cos 120pi tto f'(1)=600piapprox 1884.95559215$$
answered Nov 25 at 9:01
Mostafa Ayaz
13.7k3836
add a comment |
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$$f(x)=5sin 120pi tto f'(x)=600pi cos 120pi tto f'(1)=600piapprox 1884.95559215$$
answered Nov 25 at 9:01
Mostafa Ayaz
13.7k3836
add a comment |
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$$f(x)=5sin 120pi tto f'(x)=600pi cos 120pi tto f'(1)=600piapprox 1884.95559215$$
answered Nov 25 at 9:01
Mostafa Ayaz
13.7k3836
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$$f(x)=5sin 120pi tto f'(x)=600pi cos 120pi tto f'(1)=600piapprox 1884.95559215$$
answered Nov 25 at 9:01
Mostafa Ayaz
13.7k3836
answered Nov 25 at 9:01
Mostafa Ayaz
13.7k3836
answered Nov 25 at 9:01
Mostafa Ayaz
13.7k3836
answered Nov 25 at 9:01
Mostafa Ayaz
13.7k3836
13.7k3836
add a comment |
add a comment |
The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.
answered Nov 25 at 6:46
Andrei
10.8k21025
add a comment |
The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.
answered Nov 25 at 6:46
Andrei
10.8k21025
add a comment |
The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.
answered Nov 25 at 6:46
Andrei
10.8k21025
The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.
answered Nov 25 at 6:46
Andrei
10.8k21025
answered Nov 25 at 6:46
Andrei
10.8k21025
answered Nov 25 at 6:46
Andrei
10.8k21025
answered Nov 25 at 6:46
Andrei
10.8k21025
10.8k21025
add a comment |
add a comment |
$A = 5 sin120pi t$.
$frac{dA}{dt} = 600pi cos 120 pi t$
Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.
answered Nov 25 at 7:15
Aditya Dua
7868
add a comment |
$A = 5 sin120pi t$.
$frac{dA}{dt} = 600pi cos 120 pi t$
Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.
answered Nov 25 at 7:15
Aditya Dua
7868
add a comment |
$A = 5 sin120pi t$.
$frac{dA}{dt} = 600pi cos 120 pi t$
Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.
answered Nov 25 at 7:15
Aditya Dua
7868
$A = 5 sin120pi t$.
$frac{dA}{dt} = 600pi cos 120 pi t$
Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.
answered Nov 25 at 7:15
Aditya Dua
7868
answered Nov 25 at 7:15
Aditya Dua
7868
answered Nov 25 at 7:15
Aditya Dua
7868
answered Nov 25 at 7:15
Aditya Dua
7868
7868
add a comment |
add a comment |
$$A(t) = 5sin (120pi t)$$
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 $n$.
$$A’(1) = 600picdot 1 = 600pi$$
answered Nov 25 at 7:57
KM101
3,958417
add a comment |
$$A(t) = 5sin (120pi t)$$
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 $n$.
$$A’(1) = 600picdot 1 = 600pi$$
answered Nov 25 at 7:57
KM101
3,958417
add a comment |
$$A(t) = 5sin (120pi t)$$
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 $n$.
$$A’(1) = 600picdot 1 = 600pi$$
answered Nov 25 at 7:57
KM101
3,958417
$$A(t) = 5sin (120pi t)$$
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 $n$.
$$A’(1) = 600picdot 1 = 600pi$$
answered Nov 25 at 7:57
KM101
3,958417
answered Nov 25 at 7:57
KM101
3,958417
answered Nov 25 at 7:57
KM101
3,958417
answered Nov 25 at 7:57
KM101
3,958417
3,958417
add a comment |
add a comment |
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