increments and differentiation to determine related ratesRelated rates word problemBoyle's Law Related Rates question.Related Rates Triangle ProblemThe Chain Rule: Estimate the rate at which $θ$ is changing at that instantRelated Rates of Change rectangleCalculus Rate of Change in TrianglesAbout related ratesRelated Rates Question with right-circular cylinderCalculus Related Rates Rectangle Area ProblemRelated rates of a cube
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increments and differentiation to determine related rates
Related rates word problemBoyle's Law Related Rates question.Related Rates Triangle ProblemThe Chain Rule: Estimate the rate at which $θ$ is changing at that instantRelated Rates of Change rectangleCalculus Rate of Change in TrianglesAbout related ratesRelated Rates Question with right-circular cylinderCalculus Related Rates Rectangle Area ProblemRelated rates of a cube
$begingroup$
In an isosceles triangle with sides 20 inches long and a vertex angle of $60^circ$, the angle is closing in at $2^circ/min$. At what rate should the sides be changing to keep the area of the triangle constant (at this instant)? indicate the type of change.
calculus related-rates
edited Mar 17 at 23:09
Michael Rybkin
3,994422
asked Mar 17 at 20:56
ufotinkufotink
116
$endgroup$
add a comment |
$begingroup$
In an isosceles triangle with sides 20 inches long and a vertex angle of $60^circ$, the angle is closing in at $2^circ/min$. At what rate should the sides be changing to keep the area of the triangle constant (at this instant)? indicate the type of change.
calculus related-rates
edited Mar 17 at 23:09
Michael Rybkin
3,994422
asked Mar 17 at 20:56
ufotinkufotink
116
$endgroup$
$begingroup$
I would start with "Heron's formula" for the area of a triangle: If a triangle has side lengths a, b, and c, then the area is $sqrts(s- a)(s- b)(s- c)$ where "s" is (a+ b+ c)/2. Now add the "cosine law". If two sides of a triangle are a and b, and the angle between those sides is $theta$, then the third side is $c= sqrt{a^2+ b^2- ab cos(theta)$.
$endgroup$
– user247327
Mar 17 at 21:05
add a comment |
$begingroup$
In an isosceles triangle with sides 20 inches long and a vertex angle of $60^circ$, the angle is closing in at $2^circ/min$. At what rate should the sides be changing to keep the area of the triangle constant (at this instant)? indicate the type of change.
calculus related-rates
edited Mar 17 at 23:09
Michael Rybkin
3,994422
asked Mar 17 at 20:56
ufotinkufotink
116
$endgroup$
In an isosceles triangle with sides 20 inches long and a vertex angle of $60^circ$, the angle is closing in at $2^circ/min$. At what rate should the sides be changing to keep the area of the triangle constant (at this instant)? indicate the type of change.
calculus related-rates
calculus related-rates
edited Mar 17 at 23:09
Michael Rybkin
3,994422
asked Mar 17 at 20:56
ufotinkufotink
116
edited Mar 17 at 23:09
Michael Rybkin
3,994422
asked Mar 17 at 20:56
ufotinkufotink
116
edited Mar 17 at 23:09
Michael Rybkin
3,994422
edited Mar 17 at 23:09
Michael Rybkin
3,994422
edited Mar 17 at 23:09
Michael Rybkin
3,994422
3,994422
asked Mar 17 at 20:56
ufotinkufotink
116
asked Mar 17 at 20:56
ufotinkufotink
116
asked Mar 17 at 20:56
ufotinkufotink
116
116
$begingroup$
I would start with "Heron's formula" for the area of a triangle: If a triangle has side lengths a, b, and c, then the area is $sqrts(s- a)(s- b)(s- c)$ where "s" is (a+ b+ c)/2. Now add the "cosine law". If two sides of a triangle are a and b, and the angle between those sides is $theta$, then the third side is $c= sqrt{a^2+ b^2- ab cos(theta)$.
$endgroup$
– user247327
Mar 17 at 21:05
add a comment |
$begingroup$
I would start with "Heron's formula" for the area of a triangle: If a triangle has side lengths a, b, and c, then the area is $sqrts(s- a)(s- b)(s- c)$ where "s" is (a+ b+ c)/2. Now add the "cosine law". If two sides of a triangle are a and b, and the angle between those sides is $theta$, then the third side is $c= sqrt{a^2+ b^2- ab cos(theta)$.
$endgroup$
– user247327
Mar 17 at 21:05
$begingroup$
I would start with "Heron's formula" for the area of a triangle: If a triangle has side lengths a, b, and c, then the area is $sqrts(s- a)(s- b)(s- c)$ where "s" is (a+ b+ c)/2. Now add the "cosine law". If two sides of a triangle are a and b, and the angle between those sides is $theta$, then the third side is $c= sqrt{a^2+ b^2- ab cos(theta)$.
$endgroup$
– user247327
Mar 17 at 21:05
$begingroup$
I would start with "Heron's formula" for the area of a triangle: If a triangle has side lengths a, b, and c, then the area is $sqrts(s- a)(s- b)(s- c)$ where "s" is (a+ b+ c)/2. Now add the "cosine law". If two sides of a triangle are a and b, and the angle between those sides is $theta$, then the third side is $c= sqrt{a^2+ b^2- ab cos(theta)$.
$endgroup$
– user247327
Mar 17 at 21:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Each quantity there is a function of time including the area. $x$ is one of the two legs of the triangle. The area of the triangle: $A=hb$. Express $h$ and $b$ in terms of $x$ and $theta$:
$$costheta=frachximplies h=xcostheta\
sintheta=fracbximplies b=xsintheta$$
Therefore, we can express the area the following way:
$$A=x^2costhetasintheta=frac12x^2 2costhetasintheta=frac12x^2sin2theta$$
Saying an area is not changing with time implies that it's a constant function of time. The derivative of a constant function is zero. Implicitly differentiate both sides with respect to time and isolate $fracdxdt$—the rate of change you're looking for:
$$
fracdAdt=fracddtleft[frac12x^2sin2thetaright]implies\
0=xfracdxdtsin2theta+x^2cos2thetafracdthetadtimplies\
xfracdxdtsin2theta=-x^2cos2thetafracdthetadtimplies\
fracdxdt=-xcot2thetafracdthetadt
$$
Plug in all the known quantities at that instant, keeping in mind that $theta$ according to the picture is $30^circ$ which is equivalent to $pi/6$ in radian measure and the rate of change of the angle $theta$ should be $-1$ degree per minute or equivalently $-pi/180$ radians per minute (the negative rate of change indicates that the angle is getting smaller with time) because as the angle decreases by one degree on the right side, the same thing is happening with the other angle on the left side.
$$
fracdxdt=-20cdotcotleft(2cdotfracpi6right)cdotleft(-fracpi180right)=fracpi9cdotcotleft(fracpi3right)=fracpi9sqrt3 in/min
$$
The fact that the rate of change is positive indicates that the sides are getting larger (it's also a simple geometric observation).
answered Mar 17 at 23:07
Michael RybkinMichael Rybkin
3,994422
$endgroup$
add a comment |
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$begingroup$
Each quantity there is a function of time including the area. $x$ is one of the two legs of the triangle. The area of the triangle: $A=hb$. Express $h$ and $b$ in terms of $x$ and $theta$:
$$costheta=frachximplies h=xcostheta\
sintheta=fracbximplies b=xsintheta$$
Therefore, we can express the area the following way:
$$A=x^2costhetasintheta=frac12x^2 2costhetasintheta=frac12x^2sin2theta$$
Saying an area is not changing with time implies that it's a constant function of time. The derivative of a constant function is zero. Implicitly differentiate both sides with respect to time and isolate $fracdxdt$—the rate of change you're looking for:
$$
fracdAdt=fracddtleft[frac12x^2sin2thetaright]implies\
0=xfracdxdtsin2theta+x^2cos2thetafracdthetadtimplies\
xfracdxdtsin2theta=-x^2cos2thetafracdthetadtimplies\
fracdxdt=-xcot2thetafracdthetadt
$$
Plug in all the known quantities at that instant, keeping in mind that $theta$ according to the picture is $30^circ$ which is equivalent to $pi/6$ in radian measure and the rate of change of the angle $theta$ should be $-1$ degree per minute or equivalently $-pi/180$ radians per minute (the negative rate of change indicates that the angle is getting smaller with time) because as the angle decreases by one degree on the right side, the same thing is happening with the other angle on the left side.
$$
fracdxdt=-20cdotcotleft(2cdotfracpi6right)cdotleft(-fracpi180right)=fracpi9cdotcotleft(fracpi3right)=fracpi9sqrt3 in/min
$$
The fact that the rate of change is positive indicates that the sides are getting larger (it's also a simple geometric observation).
answered Mar 17 at 23:07
Michael RybkinMichael Rybkin
3,994422
$endgroup$
add a comment |
$begingroup$
Each quantity there is a function of time including the area. $x$ is one of the two legs of the triangle. The area of the triangle: $A=hb$. Express $h$ and $b$ in terms of $x$ and $theta$:
$$costheta=frachximplies h=xcostheta\
sintheta=fracbximplies b=xsintheta$$
Therefore, we can express the area the following way:
$$A=x^2costhetasintheta=frac12x^2 2costhetasintheta=frac12x^2sin2theta$$
Saying an area is not changing with time implies that it's a constant function of time. The derivative of a constant function is zero. Implicitly differentiate both sides with respect to time and isolate $fracdxdt$—the rate of change you're looking for:
$$
fracdAdt=fracddtleft[frac12x^2sin2thetaright]implies\
0=xfracdxdtsin2theta+x^2cos2thetafracdthetadtimplies\
xfracdxdtsin2theta=-x^2cos2thetafracdthetadtimplies\
fracdxdt=-xcot2thetafracdthetadt
$$
Plug in all the known quantities at that instant, keeping in mind that $theta$ according to the picture is $30^circ$ which is equivalent to $pi/6$ in radian measure and the rate of change of the angle $theta$ should be $-1$ degree per minute or equivalently $-pi/180$ radians per minute (the negative rate of change indicates that the angle is getting smaller with time) because as the angle decreases by one degree on the right side, the same thing is happening with the other angle on the left side.
$$
fracdxdt=-20cdotcotleft(2cdotfracpi6right)cdotleft(-fracpi180right)=fracpi9cdotcotleft(fracpi3right)=fracpi9sqrt3 in/min
$$
The fact that the rate of change is positive indicates that the sides are getting larger (it's also a simple geometric observation).
answered Mar 17 at 23:07
Michael RybkinMichael Rybkin
3,994422
$endgroup$
add a comment |
$begingroup$
Each quantity there is a function of time including the area. $x$ is one of the two legs of the triangle. The area of the triangle: $A=hb$. Express $h$ and $b$ in terms of $x$ and $theta$:
$$costheta=frachximplies h=xcostheta\
sintheta=fracbximplies b=xsintheta$$
Therefore, we can express the area the following way:
$$A=x^2costhetasintheta=frac12x^2 2costhetasintheta=frac12x^2sin2theta$$
Saying an area is not changing with time implies that it's a constant function of time. The derivative of a constant function is zero. Implicitly differentiate both sides with respect to time and isolate $fracdxdt$—the rate of change you're looking for:
$$
fracdAdt=fracddtleft[frac12x^2sin2thetaright]implies\
0=xfracdxdtsin2theta+x^2cos2thetafracdthetadtimplies\
xfracdxdtsin2theta=-x^2cos2thetafracdthetadtimplies\
fracdxdt=-xcot2thetafracdthetadt
$$
Plug in all the known quantities at that instant, keeping in mind that $theta$ according to the picture is $30^circ$ which is equivalent to $pi/6$ in radian measure and the rate of change of the angle $theta$ should be $-1$ degree per minute or equivalently $-pi/180$ radians per minute (the negative rate of change indicates that the angle is getting smaller with time) because as the angle decreases by one degree on the right side, the same thing is happening with the other angle on the left side.
$$
fracdxdt=-20cdotcotleft(2cdotfracpi6right)cdotleft(-fracpi180right)=fracpi9cdotcotleft(fracpi3right)=fracpi9sqrt3 in/min
$$
The fact that the rate of change is positive indicates that the sides are getting larger (it's also a simple geometric observation).
answered Mar 17 at 23:07
Michael RybkinMichael Rybkin
3,994422
$endgroup$
Each quantity there is a function of time including the area. $x$ is one of the two legs of the triangle. The area of the triangle: $A=hb$. Express $h$ and $b$ in terms of $x$ and $theta$:
$$costheta=frachximplies h=xcostheta\
sintheta=fracbximplies b=xsintheta$$
Therefore, we can express the area the following way:
$$A=x^2costhetasintheta=frac12x^2 2costhetasintheta=frac12x^2sin2theta$$
Saying an area is not changing with time implies that it's a constant function of time. The derivative of a constant function is zero. Implicitly differentiate both sides with respect to time and isolate $fracdxdt$—the rate of change you're looking for:
$$
fracdAdt=fracddtleft[frac12x^2sin2thetaright]implies\
0=xfracdxdtsin2theta+x^2cos2thetafracdthetadtimplies\
xfracdxdtsin2theta=-x^2cos2thetafracdthetadtimplies\
fracdxdt=-xcot2thetafracdthetadt
$$
Plug in all the known quantities at that instant, keeping in mind that $theta$ according to the picture is $30^circ$ which is equivalent to $pi/6$ in radian measure and the rate of change of the angle $theta$ should be $-1$ degree per minute or equivalently $-pi/180$ radians per minute (the negative rate of change indicates that the angle is getting smaller with time) because as the angle decreases by one degree on the right side, the same thing is happening with the other angle on the left side.
$$
fracdxdt=-20cdotcotleft(2cdotfracpi6right)cdotleft(-fracpi180right)=fracpi9cdotcotleft(fracpi3right)=fracpi9sqrt3 in/min
$$
The fact that the rate of change is positive indicates that the sides are getting larger (it's also a simple geometric observation).
answered Mar 17 at 23:07
Michael RybkinMichael Rybkin
3,994422
edited Mar 18 at 2:11
answered Mar 17 at 23:07
Michael RybkinMichael Rybkin
3,994422
answered Mar 17 at 23:07
Michael RybkinMichael Rybkin
3,994422
answered Mar 17 at 23:07
Michael RybkinMichael Rybkin
3,994422
3,994422
add a comment |
add a comment |
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$begingroup$
I would start with "Heron's formula" for the area of a triangle: If a triangle has side lengths a, b, and c, then the area is $sqrts(s- a)(s- b)(s- c)$ where "s" is (a+ b+ c)/2. Now add the "cosine law". If two sides of a triangle are a and b, and the angle between those sides is $theta$, then the third side is $c= sqrt{a^2+ b^2- ab cos(theta)$.
$endgroup$
– user247327
Mar 17 at 21:05