Introduction to related rates calculus problems:
The study of rates of change is called as calculus. Calculus related rates is dealing with finding the rates of the changes of the two variables, the two variables are related to each other in a certain way. Using the rate of change of one variable we can find the rate of change of another variable usually with respect to time.
Examples for Related Rates Calculus Problems
Q : The given problem is to solve the calculus related rates. A man and a woman start from the same point. The man walks S60o E at 1.3 m/sec. The women walks south at 1.6 m/sec. At what rate is the distance between the man and women changing after 35 minutes?
Sol : Let x be the distance the man has traveled, y the distance the women has traveled and z the distance between the man and the women.
Speed of the man, (dx/dt)=1.3 m/sec and the speed of the women, (dy/dt)=1.6 m/sec. The rate of change of the distance between them, (dz/dt) after 35 minutes have passed. After t seconds, the distance the man has traveled is 1.3t while the women’s distance is 1.6t. So after 35 minutes, or 2100 seconds, the man’s distance is x=1.3(2100)=2730m and the women’s distance y=1.6(2100)=3360m.
Use the cosine law to equate the variables.
z^2=x^2+y^2-2xycos?
The angle between the paths of the man and the women will always be 60o. Since it is a constant, substitute it into the equation before differentiating.
z^2=x^2+y^2- 2xycos(60o)
z^2=x^2+y^2 -xy
Using the chain rule, differentiate the equation with respect to time.
2z(`dz/dt` )=2x(`dx/dt` )+2y(`dy/dt` )-(y(`dx/dt` )+x(`dy/dt` ))
2z(`dz/dt` )=2x(`dx/dt` )+2y(`dy/dt` )-y(`dy/dt` )-x(`dy/dt` )
2z(`dz/dt` )=(2x-y)(`dx/dt` )+(2y-x)(`dy/dt` )
` dz/dt` = ((2x-y)(`dx/dt` )+(2y-x)(`dy/dt` ))/2z
Before substitute the information into the equation, must find the value of z after 35 minutes. x=2730 and y=3360 after 35 minutes, substitutes these values into the Cosine Law formula to find the value of z at t=35 minutes.
z^2=x^2+y^2-xy
z=sqrt(x^2+y^2-xy)
=sqrt((2730)2+(3360)2-(2730)(3360))
˜ 3093.5 m
Substitute the information given in the equation into the equation for dz/dt.
`dz/dt` = (2(2730)-(3360)(1.3))+(2(2360)-(2730)(1.6))/2(3093.5)
= ((5460-3360)(1.3)+(6720-2730)(1.6))/6187
= ((2100)(1.3)+(3990)(1.6))/6187
= (2730+6384)/6187
= `9114/6187`
˜1.47 m/sec
The calculus related rates problem are therefore 35 minutes, the distance between the man and the women is increasing at 1.47 m/sec.
Practice Problem for Related Rates Calculus Problems:
Q : The given problem is to find the calculus related rates a car leaves a gas station at an intersection at 4.30 p.m and travels east at 80 km/h. A truck traveling north at 100 km/h arrives at the same gas station at6.30 p.m. At what time are the two vehicles closest together?
Answer: Approximately 5.43.p.m
The study of rates of change is called as calculus. Calculus related rates is dealing with finding the rates of the changes of the two variables, the two variables are related to each other in a certain way. Using the rate of change of one variable we can find the rate of change of another variable usually with respect to time.
Examples for Related Rates Calculus Problems
Q : The given problem is to solve the calculus related rates. A man and a woman start from the same point. The man walks S60o E at 1.3 m/sec. The women walks south at 1.6 m/sec. At what rate is the distance between the man and women changing after 35 minutes?
Sol : Let x be the distance the man has traveled, y the distance the women has traveled and z the distance between the man and the women.
Speed of the man, (dx/dt)=1.3 m/sec and the speed of the women, (dy/dt)=1.6 m/sec. The rate of change of the distance between them, (dz/dt) after 35 minutes have passed. After t seconds, the distance the man has traveled is 1.3t while the women’s distance is 1.6t. So after 35 minutes, or 2100 seconds, the man’s distance is x=1.3(2100)=2730m and the women’s distance y=1.6(2100)=3360m.
Use the cosine law to equate the variables.
z^2=x^2+y^2-2xycos?
The angle between the paths of the man and the women will always be 60o. Since it is a constant, substitute it into the equation before differentiating.
z^2=x^2+y^2- 2xycos(60o)
z^2=x^2+y^2 -xy
Using the chain rule, differentiate the equation with respect to time.
2z(`dz/dt` )=2x(`dx/dt` )+2y(`dy/dt` )-(y(`dx/dt` )+x(`dy/dt` ))
2z(`dz/dt` )=2x(`dx/dt` )+2y(`dy/dt` )-y(`dy/dt` )-x(`dy/dt` )
2z(`dz/dt` )=(2x-y)(`dx/dt` )+(2y-x)(`dy/dt` )
` dz/dt` = ((2x-y)(`dx/dt` )+(2y-x)(`dy/dt` ))/2z
Before substitute the information into the equation, must find the value of z after 35 minutes. x=2730 and y=3360 after 35 minutes, substitutes these values into the Cosine Law formula to find the value of z at t=35 minutes.
z^2=x^2+y^2-xy
z=sqrt(x^2+y^2-xy)
=sqrt((2730)2+(3360)2-(2730)(3360))
˜ 3093.5 m
Substitute the information given in the equation into the equation for dz/dt.
`dz/dt` = (2(2730)-(3360)(1.3))+(2(2360)-(2730)(1.6))/2(3093.5)
= ((5460-3360)(1.3)+(6720-2730)(1.6))/6187
= ((2100)(1.3)+(3990)(1.6))/6187
= (2730+6384)/6187
= `9114/6187`
˜1.47 m/sec
The calculus related rates problem are therefore 35 minutes, the distance between the man and the women is increasing at 1.47 m/sec.
Practice Problem for Related Rates Calculus Problems:
Q : The given problem is to find the calculus related rates a car leaves a gas station at an intersection at 4.30 p.m and travels east at 80 km/h. A truck traveling north at 100 km/h arrives at the same gas station at6.30 p.m. At what time are the two vehicles closest together?
Answer: Approximately 5.43.p.m
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