Determine an instantaneous rate of change equation given this information
I'm in Grade 12 Advanced Functions and I've been stumped by a thinking question. Here it is.
Rael is investigating the rate of change of the function $y = cos x$ on the interval $x in [0, 2pi]$. He notices that the graph of $y = cos x$ passes through the $x$-axis at $45$ degrees. He also determines the instantaneous rate of change at $x = 0, pi$, and $2pi$ by inspection. Based on this information, determine an equation $r(x)$ to predict the instantaneous rate of change of the function $y = cos x$ on the interval $x in [0, 2pi]$. Then, use your equation to calculate the exact instantaneous rate of change at $x = pi/4$.
I know the instantaneous rate of change formula is $$frac{f(b)-f(a)}{b-a},$$ so I don't particularly see how I'm to construct my own such equation, especially implementing this information. Any help?
functions trigonometry
edited Nov 25 at 12:26
N. F. Taussig
43.4k93355
asked Nov 25 at 5:38
Korvexius
42
add a comment |
I'm in Grade 12 Advanced Functions and I've been stumped by a thinking question. Here it is.
Rael is investigating the rate of change of the function $y = cos x$ on the interval $x in [0, 2pi]$. He notices that the graph of $y = cos x$ passes through the $x$-axis at $45$ degrees. He also determines the instantaneous rate of change at $x = 0, pi$, and $2pi$ by inspection. Based on this information, determine an equation $r(x)$ to predict the instantaneous rate of change of the function $y = cos x$ on the interval $x in [0, 2pi]$. Then, use your equation to calculate the exact instantaneous rate of change at $x = pi/4$.
I know the instantaneous rate of change formula is $$frac{f(b)-f(a)}{b-a},$$ so I don't particularly see how I'm to construct my own such equation, especially implementing this information. Any help?
functions trigonometry
edited Nov 25 at 12:26
N. F. Taussig
43.4k93355
asked Nov 25 at 5:38
Korvexius
42
1
The graph crosses the $x$-axis at $90°$, not $45°$. You need to find the slope of the tangent to $cos x$ at $x = frac{pi}{4}$ by using $cos’ x$.
– KM101
Nov 25 at 5:47
This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 25 at 12:27
add a comment |
I'm in Grade 12 Advanced Functions and I've been stumped by a thinking question. Here it is.
Rael is investigating the rate of change of the function $y = cos x$ on the interval $x in [0, 2pi]$. He notices that the graph of $y = cos x$ passes through the $x$-axis at $45$ degrees. He also determines the instantaneous rate of change at $x = 0, pi$, and $2pi$ by inspection. Based on this information, determine an equation $r(x)$ to predict the instantaneous rate of change of the function $y = cos x$ on the interval $x in [0, 2pi]$. Then, use your equation to calculate the exact instantaneous rate of change at $x = pi/4$.
I know the instantaneous rate of change formula is $$frac{f(b)-f(a)}{b-a},$$ so I don't particularly see how I'm to construct my own such equation, especially implementing this information. Any help?
functions trigonometry
edited Nov 25 at 12:26
N. F. Taussig
43.4k93355
asked Nov 25 at 5:38
Korvexius
42
I'm in Grade 12 Advanced Functions and I've been stumped by a thinking question. Here it is.
Rael is investigating the rate of change of the function $y = cos x$ on the interval $x in [0, 2pi]$. He notices that the graph of $y = cos x$ passes through the $x$-axis at $45$ degrees. He also determines the instantaneous rate of change at $x = 0, pi$, and $2pi$ by inspection. Based on this information, determine an equation $r(x)$ to predict the instantaneous rate of change of the function $y = cos x$ on the interval $x in [0, 2pi]$. Then, use your equation to calculate the exact instantaneous rate of change at $x = pi/4$.
I know the instantaneous rate of change formula is $$frac{f(b)-f(a)}{b-a},$$ so I don't particularly see how I'm to construct my own such equation, especially implementing this information. Any help?
functions trigonometry
functions trigonometry
edited Nov 25 at 12:26
N. F. Taussig
43.4k93355
asked Nov 25 at 5:38
Korvexius
42
edited Nov 25 at 12:26
N. F. Taussig
43.4k93355
asked Nov 25 at 5:38
Korvexius
42
edited Nov 25 at 12:26
N. F. Taussig
43.4k93355
edited Nov 25 at 12:26
N. F. Taussig
43.4k93355
edited Nov 25 at 12:26
N. F. Taussig
43.4k93355
43.4k93355
asked Nov 25 at 5:38
Korvexius
42
asked Nov 25 at 5:38
Korvexius
42
asked Nov 25 at 5:38
Korvexius
42
42
1
The graph crosses the $x$-axis at $90°$, not $45°$. You need to find the slope of the tangent to $cos x$ at $x = frac{pi}{4}$ by using $cos’ x$.
– KM101
Nov 25 at 5:47
This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 25 at 12:27
add a comment |
1
The graph crosses the $x$-axis at $90°$, not $45°$. You need to find the slope of the tangent to $cos x$ at $x = frac{pi}{4}$ by using $cos’ x$.
– KM101
Nov 25 at 5:47
This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 25 at 12:27
1
1
The graph crosses the $x$-axis at $90°$, not $45°$. You need to find the slope of the tangent to $cos x$ at $x = frac{pi}{4}$ by using $cos’ x$.
– KM101
Nov 25 at 5:47
The graph crosses the $x$-axis at $90°$, not $45°$. You need to find the slope of the tangent to $cos x$ at $x = frac{pi}{4}$ by using $cos’ x$.
– KM101
Nov 25 at 5:47
This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 25 at 12:27
This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 25 at 12:27
add a comment |
2 Answers
2
active
oldest
votes
There isn’t any “information” given by the question, so you have to find the equation for $cos’ x$ (the derivative of $cos x$). Usually, the following formulas are used.
$$f’(x) = lim_{h to 0}frac{f(x+h)-f(x)}{h} = lim_{b to x}frac{f(b)-f(x)}{b-x}$$
Which are both equivalent (if you let $h = b-x$).
If you know about derivatives, you’ll know that $cos’x = -sin x$, which gives the instantaneous rate of change. If not, here’s one way to derive it.
$$lim_{h to 0}frac{cos(x+h)-cos x}{h}$$
$$= lim_{h to 0}frac{color{blue}{cos xcos h}-sin xsin hcolor{blue}{-cos x}}{h}$$
$$= lim_{h to 0}frac{color{blue}{cos x(cos h-1)}-sin xsin h}{h}$$
$$= lim_{h to 0}frac{cos x(cos h-1)}{h}-lim_{h to 0}frac{sin xsin h}{h}$$
$$= cos xcdotlim_{h to 0}frac{cos h-1}{h}-sin xcdotlim_{h to 0}frac{sin h}{h}$$
Using $lim_limits{h to 0} frac{sin h}{h} = 1$ and $lim_limits{h to 0}frac{cos h-1}{h} = 0$, you get
$$= cos xcdot 0 - sin xcdot 1 = color{purple}{-sin x}$$
The instantaneous slope/slope of the tangent at point $x_1$ can be found by $$m = f’(x_1) implies m = -sin(x_1)$$
The given point is $x_1 = frac{pi}{4}$, so you get
$$m = -sinbigg(frac{pi}{4}bigg) = -frac{sqrt 2}{2}$$
answered Nov 25 at 6:17
KM101
3,958417
add a comment |
Not sure how much calculus do you know but here is why differentiating cosine gives you negative of sine.
begin{align}
lim_{b to a} frac{cos(b)-cos(a)}{b-a} &=lim_{b to a} frac{-2 sinleft( frac{a+b}2right) sinleft(frac{b-a}2 right)}{b-a} \
&=lim_{b to a}- sinleft( frac{a+b}2right)lim_{b to a} frac{ sinleft(frac{b-a}2 right)}{frac{b-a}2} \
&=lim_{b to a}- sinleft( frac{a+b}2right)\
&= - sin (a)
end{align}
answered Nov 25 at 5:51
Siong Thye Goh
98.4k1463116
Calculus is next semester for me, sir.
– Korvexius
Nov 25 at 6:08
Then I am not sure how can you answer the question.
– Siong Thye Goh
Nov 25 at 7:19
Ah, your answer is right. Looks like it was just a really, really hard question for my class.
– Korvexius
Nov 29 at 4:49
add a comment |
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There isn’t any “information” given by the question, so you have to find the equation for $cos’ x$ (the derivative of $cos x$). Usually, the following formulas are used.
$$f’(x) = lim_{h to 0}frac{f(x+h)-f(x)}{h} = lim_{b to x}frac{f(b)-f(x)}{b-x}$$
Which are both equivalent (if you let $h = b-x$).
If you know about derivatives, you’ll know that $cos’x = -sin x$, which gives the instantaneous rate of change. If not, here’s one way to derive it.
$$lim_{h to 0}frac{cos(x+h)-cos x}{h}$$
$$= lim_{h to 0}frac{color{blue}{cos xcos h}-sin xsin hcolor{blue}{-cos x}}{h}$$
$$= lim_{h to 0}frac{color{blue}{cos x(cos h-1)}-sin xsin h}{h}$$
$$= lim_{h to 0}frac{cos x(cos h-1)}{h}-lim_{h to 0}frac{sin xsin h}{h}$$
$$= cos xcdotlim_{h to 0}frac{cos h-1}{h}-sin xcdotlim_{h to 0}frac{sin h}{h}$$
Using $lim_limits{h to 0} frac{sin h}{h} = 1$ and $lim_limits{h to 0}frac{cos h-1}{h} = 0$, you get
$$= cos xcdot 0 - sin xcdot 1 = color{purple}{-sin x}$$
The instantaneous slope/slope of the tangent at point $x_1$ can be found by $$m = f’(x_1) implies m = -sin(x_1)$$
The given point is $x_1 = frac{pi}{4}$, so you get
$$m = -sinbigg(frac{pi}{4}bigg) = -frac{sqrt 2}{2}$$
answered Nov 25 at 6:17
KM101
3,958417
add a comment |
There isn’t any “information” given by the question, so you have to find the equation for $cos’ x$ (the derivative of $cos x$). Usually, the following formulas are used.
$$f’(x) = lim_{h to 0}frac{f(x+h)-f(x)}{h} = lim_{b to x}frac{f(b)-f(x)}{b-x}$$
Which are both equivalent (if you let $h = b-x$).
If you know about derivatives, you’ll know that $cos’x = -sin x$, which gives the instantaneous rate of change. If not, here’s one way to derive it.
$$lim_{h to 0}frac{cos(x+h)-cos x}{h}$$
$$= lim_{h to 0}frac{color{blue}{cos xcos h}-sin xsin hcolor{blue}{-cos x}}{h}$$
$$= lim_{h to 0}frac{color{blue}{cos x(cos h-1)}-sin xsin h}{h}$$
$$= lim_{h to 0}frac{cos x(cos h-1)}{h}-lim_{h to 0}frac{sin xsin h}{h}$$
$$= cos xcdotlim_{h to 0}frac{cos h-1}{h}-sin xcdotlim_{h to 0}frac{sin h}{h}$$
Using $lim_limits{h to 0} frac{sin h}{h} = 1$ and $lim_limits{h to 0}frac{cos h-1}{h} = 0$, you get
$$= cos xcdot 0 - sin xcdot 1 = color{purple}{-sin x}$$
The instantaneous slope/slope of the tangent at point $x_1$ can be found by $$m = f’(x_1) implies m = -sin(x_1)$$
The given point is $x_1 = frac{pi}{4}$, so you get
$$m = -sinbigg(frac{pi}{4}bigg) = -frac{sqrt 2}{2}$$
answered Nov 25 at 6:17
KM101
3,958417
add a comment |
There isn’t any “information” given by the question, so you have to find the equation for $cos’ x$ (the derivative of $cos x$). Usually, the following formulas are used.
$$f’(x) = lim_{h to 0}frac{f(x+h)-f(x)}{h} = lim_{b to x}frac{f(b)-f(x)}{b-x}$$
Which are both equivalent (if you let $h = b-x$).
If you know about derivatives, you’ll know that $cos’x = -sin x$, which gives the instantaneous rate of change. If not, here’s one way to derive it.
$$lim_{h to 0}frac{cos(x+h)-cos x}{h}$$
$$= lim_{h to 0}frac{color{blue}{cos xcos h}-sin xsin hcolor{blue}{-cos x}}{h}$$
$$= lim_{h to 0}frac{color{blue}{cos x(cos h-1)}-sin xsin h}{h}$$
$$= lim_{h to 0}frac{cos x(cos h-1)}{h}-lim_{h to 0}frac{sin xsin h}{h}$$
$$= cos xcdotlim_{h to 0}frac{cos h-1}{h}-sin xcdotlim_{h to 0}frac{sin h}{h}$$
Using $lim_limits{h to 0} frac{sin h}{h} = 1$ and $lim_limits{h to 0}frac{cos h-1}{h} = 0$, you get
$$= cos xcdot 0 - sin xcdot 1 = color{purple}{-sin x}$$
The instantaneous slope/slope of the tangent at point $x_1$ can be found by $$m = f’(x_1) implies m = -sin(x_1)$$
The given point is $x_1 = frac{pi}{4}$, so you get
$$m = -sinbigg(frac{pi}{4}bigg) = -frac{sqrt 2}{2}$$
answered Nov 25 at 6:17
KM101
3,958417
There isn’t any “information” given by the question, so you have to find the equation for $cos’ x$ (the derivative of $cos x$). Usually, the following formulas are used.
$$f’(x) = lim_{h to 0}frac{f(x+h)-f(x)}{h} = lim_{b to x}frac{f(b)-f(x)}{b-x}$$
Which are both equivalent (if you let $h = b-x$).
If you know about derivatives, you’ll know that $cos’x = -sin x$, which gives the instantaneous rate of change. If not, here’s one way to derive it.
$$lim_{h to 0}frac{cos(x+h)-cos x}{h}$$
$$= lim_{h to 0}frac{color{blue}{cos xcos h}-sin xsin hcolor{blue}{-cos x}}{h}$$
$$= lim_{h to 0}frac{color{blue}{cos x(cos h-1)}-sin xsin h}{h}$$
$$= lim_{h to 0}frac{cos x(cos h-1)}{h}-lim_{h to 0}frac{sin xsin h}{h}$$
$$= cos xcdotlim_{h to 0}frac{cos h-1}{h}-sin xcdotlim_{h to 0}frac{sin h}{h}$$
Using $lim_limits{h to 0} frac{sin h}{h} = 1$ and $lim_limits{h to 0}frac{cos h-1}{h} = 0$, you get
$$= cos xcdot 0 - sin xcdot 1 = color{purple}{-sin x}$$
The instantaneous slope/slope of the tangent at point $x_1$ can be found by $$m = f’(x_1) implies m = -sin(x_1)$$
The given point is $x_1 = frac{pi}{4}$, so you get
$$m = -sinbigg(frac{pi}{4}bigg) = -frac{sqrt 2}{2}$$
answered Nov 25 at 6:17
KM101
3,958417
edited Nov 25 at 6:22
answered Nov 25 at 6:17
KM101
3,958417
answered Nov 25 at 6:17
KM101
3,958417
answered Nov 25 at 6:17
KM101
3,958417
3,958417
add a comment |
add a comment |
Not sure how much calculus do you know but here is why differentiating cosine gives you negative of sine.
begin{align}
lim_{b to a} frac{cos(b)-cos(a)}{b-a} &=lim_{b to a} frac{-2 sinleft( frac{a+b}2right) sinleft(frac{b-a}2 right)}{b-a} \
&=lim_{b to a}- sinleft( frac{a+b}2right)lim_{b to a} frac{ sinleft(frac{b-a}2 right)}{frac{b-a}2} \
&=lim_{b to a}- sinleft( frac{a+b}2right)\
&= - sin (a)
end{align}
answered Nov 25 at 5:51
Siong Thye Goh
98.4k1463116
Calculus is next semester for me, sir.
– Korvexius
Nov 25 at 6:08
Then I am not sure how can you answer the question.
– Siong Thye Goh
Nov 25 at 7:19
Ah, your answer is right. Looks like it was just a really, really hard question for my class.
– Korvexius
Nov 29 at 4:49
add a comment |
Not sure how much calculus do you know but here is why differentiating cosine gives you negative of sine.
begin{align}
lim_{b to a} frac{cos(b)-cos(a)}{b-a} &=lim_{b to a} frac{-2 sinleft( frac{a+b}2right) sinleft(frac{b-a}2 right)}{b-a} \
&=lim_{b to a}- sinleft( frac{a+b}2right)lim_{b to a} frac{ sinleft(frac{b-a}2 right)}{frac{b-a}2} \
&=lim_{b to a}- sinleft( frac{a+b}2right)\
&= - sin (a)
end{align}
answered Nov 25 at 5:51
Siong Thye Goh
98.4k1463116
Calculus is next semester for me, sir.
– Korvexius
Nov 25 at 6:08
Then I am not sure how can you answer the question.
– Siong Thye Goh
Nov 25 at 7:19
Ah, your answer is right. Looks like it was just a really, really hard question for my class.
– Korvexius
Nov 29 at 4:49
add a comment |
Not sure how much calculus do you know but here is why differentiating cosine gives you negative of sine.
begin{align}
lim_{b to a} frac{cos(b)-cos(a)}{b-a} &=lim_{b to a} frac{-2 sinleft( frac{a+b}2right) sinleft(frac{b-a}2 right)}{b-a} \
&=lim_{b to a}- sinleft( frac{a+b}2right)lim_{b to a} frac{ sinleft(frac{b-a}2 right)}{frac{b-a}2} \
&=lim_{b to a}- sinleft( frac{a+b}2right)\
&= - sin (a)
end{align}
answered Nov 25 at 5:51
Siong Thye Goh
98.4k1463116
Not sure how much calculus do you know but here is why differentiating cosine gives you negative of sine.
begin{align}
lim_{b to a} frac{cos(b)-cos(a)}{b-a} &=lim_{b to a} frac{-2 sinleft( frac{a+b}2right) sinleft(frac{b-a}2 right)}{b-a} \
&=lim_{b to a}- sinleft( frac{a+b}2right)lim_{b to a} frac{ sinleft(frac{b-a}2 right)}{frac{b-a}2} \
&=lim_{b to a}- sinleft( frac{a+b}2right)\
&= - sin (a)
end{align}
answered Nov 25 at 5:51
Siong Thye Goh
98.4k1463116
answered Nov 25 at 5:51
Siong Thye Goh
98.4k1463116
answered Nov 25 at 5:51
Siong Thye Goh
98.4k1463116
answered Nov 25 at 5:51
Siong Thye Goh
98.4k1463116
98.4k1463116
Calculus is next semester for me, sir.
– Korvexius
Nov 25 at 6:08
Then I am not sure how can you answer the question.
– Siong Thye Goh
Nov 25 at 7:19
Ah, your answer is right. Looks like it was just a really, really hard question for my class.
– Korvexius
Nov 29 at 4:49
add a comment |
Calculus is next semester for me, sir.
– Korvexius
Nov 25 at 6:08
Then I am not sure how can you answer the question.
– Siong Thye Goh
Nov 25 at 7:19
Ah, your answer is right. Looks like it was just a really, really hard question for my class.
– Korvexius
Nov 29 at 4:49
Calculus is next semester for me, sir.
– Korvexius
Nov 25 at 6:08
Calculus is next semester for me, sir.
– Korvexius
Nov 25 at 6:08
Then I am not sure how can you answer the question.
– Siong Thye Goh
Nov 25 at 7:19
Then I am not sure how can you answer the question.
– Siong Thye Goh
Nov 25 at 7:19
Ah, your answer is right. Looks like it was just a really, really hard question for my class.
– Korvexius
Nov 29 at 4:49
Ah, your answer is right. Looks like it was just a really, really hard question for my class.
– Korvexius
Nov 29 at 4:49
add a comment |
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1
The graph crosses the $x$-axis at $90°$, not $45°$. You need to find the slope of the tangent to $cos x$ at $x = frac{pi}{4}$ by using $cos’ x$.
– KM101
Nov 25 at 5:47
This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 25 at 12:27