Mixing
related rates calculus question
$begingroup$
A balloon is at a height of 20 meters, and is rising at the constant rate of 5 m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10 m/sec. How fast is the distance between the bicyclist and the balloon increasing 2 seconds later?
calculus
asked Oct 22 '15 at 18:43
Valentina VacaValentina Vaca
175
$endgroup$
add a comment |
$begingroup$
A balloon is at a height of 20 meters, and is rising at the constant rate of 5 m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10 m/sec. How fast is the distance between the bicyclist and the balloon increasing 2 seconds later?
calculus
asked Oct 22 '15 at 18:43
Valentina VacaValentina Vaca
175
$endgroup$
add a comment |
$begingroup$
A balloon is at a height of 20 meters, and is rising at the constant rate of 5 m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10 m/sec. How fast is the distance between the bicyclist and the balloon increasing 2 seconds later?
calculus
asked Oct 22 '15 at 18:43
Valentina VacaValentina Vaca
175
$endgroup$
A balloon is at a height of 20 meters, and is rising at the constant rate of 5 m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10 m/sec. How fast is the distance between the bicyclist and the balloon increasing 2 seconds later?
calculus
calculus
asked Oct 22 '15 at 18:43
Valentina VacaValentina Vaca
175
asked Oct 22 '15 at 18:43
Valentina VacaValentina Vaca
175
asked Oct 22 '15 at 18:43
Valentina VacaValentina Vaca
175
asked Oct 22 '15 at 18:43
Valentina VacaValentina Vaca
175
asked Oct 22 '15 at 18:43
Valentina VacaValentina Vaca
175
175
add a comment |
add a comment |
1 Answer
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$begingroup$
Consider a right angled triangle where the height l represents the balloon's height, the base b is the bicycle's travelled distance and hypotenuse h is the required distance between balloon and bicycle.
So we have by Pythagoras theorem, $l^2+b^2=h^2$ or, $lfrac{dl}{dt}+bfrac{db}{dt}=hfrac{dh}{dt}$
Now put the values and do the rest of the work. If you face any problem, I will do it for you.
answered Oct 22 '15 at 18:53
SchrodingersCatSchrodingersCat
22.4k52862
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider a right angled triangle where the height l represents the balloon's height, the base b is the bicycle's travelled distance and hypotenuse h is the required distance between balloon and bicycle.
So we have by Pythagoras theorem, $l^2+b^2=h^2$ or, $lfrac{dl}{dt}+bfrac{db}{dt}=hfrac{dh}{dt}$
Now put the values and do the rest of the work. If you face any problem, I will do it for you.
answered Oct 22 '15 at 18:53
SchrodingersCatSchrodingersCat
22.4k52862
$endgroup$
add a comment |
$begingroup$
Consider a right angled triangle where the height l represents the balloon's height, the base b is the bicycle's travelled distance and hypotenuse h is the required distance between balloon and bicycle.
So we have by Pythagoras theorem, $l^2+b^2=h^2$ or, $lfrac{dl}{dt}+bfrac{db}{dt}=hfrac{dh}{dt}$
Now put the values and do the rest of the work. If you face any problem, I will do it for you.
answered Oct 22 '15 at 18:53
SchrodingersCatSchrodingersCat
22.4k52862
$endgroup$
add a comment |
$begingroup$
Consider a right angled triangle where the height l represents the balloon's height, the base b is the bicycle's travelled distance and hypotenuse h is the required distance between balloon and bicycle.
So we have by Pythagoras theorem, $l^2+b^2=h^2$ or, $lfrac{dl}{dt}+bfrac{db}{dt}=hfrac{dh}{dt}$
Now put the values and do the rest of the work. If you face any problem, I will do it for you.
answered Oct 22 '15 at 18:53
SchrodingersCatSchrodingersCat
22.4k52862
$endgroup$
Consider a right angled triangle where the height l represents the balloon's height, the base b is the bicycle's travelled distance and hypotenuse h is the required distance between balloon and bicycle.
So we have by Pythagoras theorem, $l^2+b^2=h^2$ or, $lfrac{dl}{dt}+bfrac{db}{dt}=hfrac{dh}{dt}$
Now put the values and do the rest of the work. If you face any problem, I will do it for you.
answered Oct 22 '15 at 18:53
SchrodingersCatSchrodingersCat
22.4k52862
answered Oct 22 '15 at 18:53
SchrodingersCatSchrodingersCat
22.4k52862
answered Oct 22 '15 at 18:53
SchrodingersCatSchrodingersCat
22.4k52862
answered Oct 22 '15 at 18:53
SchrodingersCatSchrodingersCat
22.4k52862
22.4k52862
add a comment |
add a comment |
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