An airplane flies at an altitude of 5 miles toward a point directly over an observer (see figure). The speed of the plane is 600 miles per hour. Find the rates at which the angle of elevation&nbsp;θ&nbsp;is changing when the angle is&nbsp;a)&nbsp;θ=30∘&nbsp;b)&nbsp;θ=75∘

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An airplane flies at an altitude of 5 miles toward a point directly over an observer (see figure). The speed of the plane is 600 miles per hour. Find the rates at which the angle of elevation&amp;nbsp;θ&amp;nbsp;is changing when the angle is&amp;nbsp;a)&amp;nbsp;θ=30∘&amp;nbsp;b)&amp;nbsp;θ=75∘

Feas1981 · Accepted Answer

Linking and theta.  Speed in x direction, which is equal to the derivative of x with respects to time, is given  tam(θ)=5mix  dxdt=600mih  Take derivative of function  ddt(tan⁡θ=5mix)  sec2(θ)dθdt=−5mix2⋅dxdt  Solve for dθdt  dθdt=−5misec2(θ)⋅x2⋅dxdt  dθdt=−5mi⋅cos2(θ)x2⋅−600mih  Solve original function for x  tan⁡(θ)=5mix  x=5mitan⁡(θ)  Plug in x and simplify  dθdt=−5mi⋅cos2(θ)(5mitan⁡(θ))2⋅−600mih  dθdt=−5mi⋅cos2(θ)25mi2⋅cos⁡(θ)2⋅−600mih  dθdt=−1mi⋅sin2(θ)5mi2⋅−600mih  dθdt=120⋅sin2(θ)h−1  Plug in θ values  a) θ=30∘

Jeffrey Jordon · Accepted Answer

We can solve this problem using trigonometry and calculus. Let&#039;s define some variables:- h: the altitude of the plane (in miles)- d: the horizontal distance between the plane and the observer (in miles)- theta: the angle of elevation (in radians)- phi: the angle between the plane&#039;s path and the horizontal (in radians)We can use the tangent function to relate the variables:tan(θ)=hdTaking the derivative of both sides with respect to time t, we get:sec2(θ)dθdt=1ddhdt−hd2dddtWe can also use the cosine function to relate phi and theta:cos(ϕ)=dh2+d2Taking the derivative of both sides with respect to time t, we get:−sin(ϕ)dϕdt=1h2+d2dddt−h(h2+d2)3/2dhdtNow we can substitute the given values and solve for the rates of change:a) When θ=30∘, we have tan(θ)=hd=h5=33.Solving for h and d, we get h=533 miles and d=53 miles.Since the plane&#039;s speed is 600 miles per hour, we have dddt=−600sin(ϕ).Using the cosine function, we can find ϕ:cos(ϕ)=dh2+d2=32Therefore, ϕ=30∘.Using the equations above, we can find the rates of change:dθdt=1sec2(θ)(1ddhdt−hd2dddt)=−1203269≈−1.99 radians per hour.b) When θ=75∘, we have tan(θ)=hd=h5=3.732.Solving for h and d, we get h≈18.66 miles and d≈4.999 miles.Using the cosine function, we can find ϕ:cos(ϕ)=dh2+d2≈0.267Therefore, ϕ≈74.05∘.Using the equations above, we can find the rates of change:dθdt=1sec2(θ)(1ddhdt−hd2dddt)≈0.038 radians per hour

alenahelenash · Accepted Answer

Result:30∘ and 75∘Solution:Given:tan(θ)=5xTo find the rates at which the angle of elevation θ is changing with respect to time, we need to differentiate both sides of this equation with respect to time. Let&#039;s denote the rate at which θ is changing as dθdt and the rate at which x is changing as dxdt. Applying the chain rule, we get:sec2(θ)dθdt=−5x2dxdtNow, let&#039;s solve for dθdt.a) When θ=30∘:Substituting θ=30∘ into the equation above, we have:sec2(30∘)dθdt=−5x2dxdtSince sec(30∘)=23, the equation becomes:(23)2dθdt=−5x2dxdtSimplifying, we get:43dθdt=−5x2dxdtTo find dθdt, we divide both sides by 43:dθdt=−54x2dxdtb) When θ=75∘:Substituting θ=75∘ into the original equation, we have:sec2(75∘)dθdt=−5x2dxdtSince sec(75∘)=6+24, the equation becomes:(6+24)2dθdt=−5x2dxdtSimplifying, we get:8+4316dθdt=−5x2dxdtTo find dθdt, we divide both sides by 8+4316:dθdt=−5x2dxdt·168+43Simplifying the expression on the right-hand side, we get:dθdt=−80(x2)(8+43)dxdtThese are the rates at which the angle of elevation θ is changing when θ is equal to 30∘ and 75∘, respectively.

star233 · Accepted Answer

We can use trigonometry to relate the quantities x, θ, and the altitude of the airplane. The tangent function is useful in this case:tan(θ)=altitudedistanceSubstituting the given values, we have:tan(θ)=5xNow, we can differentiate both sides of the equation with respect to time t to find the rates at which the angle of elevation is changing.a) When θ=30 degrees:Differentiating both sides with respect to t gives:sec2(θ)·dθdt=−5x2·dxdtSince θ=30 degrees, we have sec2(θ)=43:43·dθdt=−5x2·dxdtTo find the rates at which θ is changing, we need to know the rate at which x is changing. Unfortunately, this information is not provided in the problem statement. Without it, we cannot determine the rate at which θ is changing when θ=30 degrees.b) When θ=75 degrees:Again, differentiating both sides with respect to t gives:sec2(θ)·dθdt=−5x2·dxdtSince θ=75 degrees, we have sec2(θ)=4:4·dθdt=−5x2·dxdtTo find the rates at which θ is changing, we need to know the rate at which x is changing. Without this information, we cannot determine the rate at which θ is changing when θ=75 degrees.In conclusion, without knowing the rate at which the distance x changes, we cannot determine the rates at which the angle of elevation θ is changing for the given values of θ.

An airplane flies at an altitude of 5 miles toward a point directly over an obse

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