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A road perpendicular to a highway leads to a farmhouse located 2 mile away. An automobile traveling on the highway passes through this intersection at a speed of 75mph. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 4 miles past the intersection of the highway and the road? The distance between the automobile and the farmhouse is increasing at a rate of miles per hour.

Respuesta :

Answer:

Hence the distance between automobile and farmhouse is increase at rate of 33.541 mph.

Step-by-step explanation:

Given:

farmhouse distance from highway is 2 miles

An automobile traveling at 75 mph.

To Find:

Distance between automobile and farmhouse when automobile pass 4 miles way from intersection of highway.(rate miles per hr)

Solution:

The solution is required in rate ,(Refer the attachment)

so using derivative with respect to time  

Consider Triangle ABC,

AB=2 miles

x=BC=X km distance from highway intersection

y=AC= Distance between car and the farmhouse.

Using Pythagoras Theorem we get ,

y^2=x^2+2^2

y^2=x^2+4

y=Sqrt of{(x^2+4)}

Differentiate w.r.t 't' we get

2y*(dy/dt)=2x*(dx/dt)

dy/dt=(x/y)*(dx/dt)

dy/dt=(x/Sqrt of{(x^2+4)})*(dx/dt)............(here x=4 miles)

dy/dt=(2/Sqrt(20))*(dx/dt)

here dx/dt is the rate of change of distance i.e. speed=75

dy/dt=(2/Sqrt(20))*(75)

=0.44721*75

=33.541 mph

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ajeigbeibraheem

A road perpendicular to a highway leads to a farmhouse located 2 miles away. An automobile traveling on the highway passes through this intersection at a speed of 75mph. The distance of how fast the automobile to the farmhouse is p = 4.47 miles. The distance between the automobile and the farmhouse is increasing at the rate of  67.11 miles

From the given information:

  • the perpendicular distance of the road to the high = 2 miles
  • Let the distance between the automobile and the farmhouse = p miles
  • the distance between the intersection and the automobile be = q miles

Thus, to find the rate at which p(miles) is increasing, when q = 4 miles is determined as follows:

  • p² = 2² + q²

By differentiation;

[tex]\mathbf{2p\times \dfrac{dp}{dt} = 0 + 2q \times \dfrac{dq}{dt} }[/tex]

[tex]\mathbf{\dfrac{dp}{dt} = (\dfrac{q}{s}) (\dfrac{dq}{dt})}[/tex]

Recall that:

[tex]\mathbf{\dfrac{dq}{dt} = 75 \ mph}[/tex] and q = 4 miles.

∴

p² = 2² + 4²

p² = 4 + 16

p² = 20

[tex]\mathbf{p = \sqrt{20}}[/tex]

p = 4.47 miles

Now; the distance between the automobile and the farmhouse is increasing at the rate of [tex]\mathbf{\dfrac{dp}{dt} = (\dfrac{q}{s}) (\dfrac{dq}{dt})}[/tex]

[tex]\mathbf{\dfrac{dp}{dt} = (\dfrac{4}{4.47}) \times 75}[/tex]

[tex]\mathbf{\dfrac{dp}{dt} =67.11 \ miles}[/tex]

Learn more about distance here:

https://brainly.com/question/12319416?referrer=searchResults

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