$A = 5sin120pi t$, find rate of change at $t = 1s$.
$begingroup$
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 '18 at 7:58
jayant98
501116
asked Nov 25 '18 at 6:06
KorvexiusKorvexius
72
$endgroup$
add a comment |
$begingroup$
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 '18 at 7:58
jayant98
501116
asked Nov 25 '18 at 6:06
KorvexiusKorvexius
72
$endgroup$
add a comment |
$begingroup$
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 '18 at 7:58
jayant98
501116
asked Nov 25 '18 at 6:06
KorvexiusKorvexius
72
$endgroup$
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 '18 at 7:58
jayant98
501116
asked Nov 25 '18 at 6:06
KorvexiusKorvexius
72
edited Nov 25 '18 at 7:58
jayant98
501116
asked Nov 25 '18 at 6:06
KorvexiusKorvexius
72
edited Nov 25 '18 at 7:58
jayant98
501116
edited Nov 25 '18 at 7:58
jayant98
501116
edited Nov 25 '18 at 7:58
jayant98
501116
501116
asked Nov 25 '18 at 6:06
KorvexiusKorvexius
72
asked Nov 25 '18 at 6:06
KorvexiusKorvexius
72
asked Nov 25 '18 at 6:06
KorvexiusKorvexius
72
72
add a comment |
add a comment |
4 Answers
4
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oldest
votes
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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 '18 at 9:01
Mostafa AyazMostafa Ayaz
15.2k3939
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$begingroup$
The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.
answered Nov 25 '18 at 6:46
AndreiAndrei
11.7k21026
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$begingroup$
$A = 5 sin120pi t$.
$frac{dA}{dt} = 600pi cos 120 pi t$
Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.
answered Nov 25 '18 at 7:15
Aditya DuaAditya Dua
95918
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add a comment |
$begingroup$
$$A(t) = 5sin (120pi t)$$
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 '18 at 7:57
KM101KM101
5,9661523
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add a comment |
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4 Answers
4
active
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votes
4 Answers
4
active
oldest
votes
active
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active
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votes
$begingroup$
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 '18 at 9:01
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
add a comment |
$begingroup$
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 '18 at 9:01
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
add a comment |
$begingroup$
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 '18 at 9:01
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
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 '18 at 9:01
Mostafa AyazMostafa Ayaz
15.2k3939
answered Nov 25 '18 at 9:01
Mostafa AyazMostafa Ayaz
15.2k3939
answered Nov 25 '18 at 9:01
Mostafa AyazMostafa Ayaz
15.2k3939
answered Nov 25 '18 at 9:01
Mostafa AyazMostafa Ayaz
15.2k3939
15.2k3939
add a comment |
add a comment |
$begingroup$
The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.
answered Nov 25 '18 at 6:46
AndreiAndrei
11.7k21026
$endgroup$
add a comment |
$begingroup$
The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.
answered Nov 25 '18 at 6:46
AndreiAndrei
11.7k21026
$endgroup$
add a comment |
$begingroup$
The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.
answered Nov 25 '18 at 6:46
AndreiAndrei
11.7k21026
$endgroup$
The current changing rate is $frac{dA}{dt}$. Just take the derivative, then plug in $t=1$.
answered Nov 25 '18 at 6:46
AndreiAndrei
11.7k21026
answered Nov 25 '18 at 6:46
AndreiAndrei
11.7k21026
answered Nov 25 '18 at 6:46
AndreiAndrei
11.7k21026
answered Nov 25 '18 at 6:46
AndreiAndrei
11.7k21026
11.7k21026
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add a comment |
$begingroup$
$A = 5 sin120pi t$.
$frac{dA}{dt} = 600pi cos 120 pi t$
Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.
answered Nov 25 '18 at 7:15
Aditya DuaAditya Dua
95918
$endgroup$
add a comment |
$begingroup$
$A = 5 sin120pi t$.
$frac{dA}{dt} = 600pi cos 120 pi t$
Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.
answered Nov 25 '18 at 7:15
Aditya DuaAditya Dua
95918
$endgroup$
add a comment |
$begingroup$
$A = 5 sin120pi t$.
$frac{dA}{dt} = 600pi cos 120 pi t$
Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.
answered Nov 25 '18 at 7:15
Aditya DuaAditya Dua
95918
$endgroup$
$A = 5 sin120pi t$.
$frac{dA}{dt} = 600pi cos 120 pi t$
Evaluated at $t=1s$, the rate of change of current is $600pi cos120pi = 600 pi ;A/s$.
answered Nov 25 '18 at 7:15
Aditya DuaAditya Dua
95918
answered Nov 25 '18 at 7:15
Aditya DuaAditya Dua
95918
answered Nov 25 '18 at 7:15
Aditya DuaAditya Dua
95918
answered Nov 25 '18 at 7:15
Aditya DuaAditya Dua
95918
95918
add a comment |
add a comment |
$begingroup$
$$A(t) = 5sin (120pi t)$$
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 '18 at 7:57
KM101KM101
5,9661523
$endgroup$
add a comment |
$begingroup$
$$A(t) = 5sin (120pi t)$$
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 '18 at 7:57
KM101KM101
5,9661523
$endgroup$
add a comment |
$begingroup$
$$A(t) = 5sin (120pi t)$$
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 '18 at 7:57
KM101KM101
5,9661523
$endgroup$
$$A(t) = 5sin (120pi t)$$
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 '18 at 7:57
KM101KM101
5,9661523
answered Nov 25 '18 at 7:57
KM101KM101
5,9661523
answered Nov 25 '18 at 7:57
KM101KM101
5,9661523
answered Nov 25 '18 at 7:57
KM101KM101
5,9661523
5,9661523
add a comment |
add a comment |
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