Calculus HW giving me a lot of trouble...Please help!

R

rofl

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3. A plane flying with a constant speed of 14 km/min passes over a ground radar station at an altitude of 8 km and climbs at an angle of 50 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 1 minutes later?




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5. A particle is moving along the curve y=2*sqrt(5x+1). As the particle passes through the point (38), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.
 
What have you tried to do with these problems?
Do you simply want then done for you?
There are sites that will let you pay to have them done for you.
This is not one of those.
 
1. At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 25 knots and ship B is sailing north at 18 knots. How fast (in knots) is the distance between the ships changing at 7 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
Click to expand...


Anytime you see a problem like this, think Pythagoras. Initially, Ship A is 50 miles from Ship B. Ask yourself, after 7 hours how far apart are the ships?. d=rt. You know the respective rates and times. Now, use those to find D, the distance between the ships after 7 hours. You can use \(\displaystyle \L\\D=x^{2}+y^{2}\) to find the rate of change by differentiating: \(\displaystyle \L\\2D\frac{dD}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\). Fill in your info and solve for dD/dt.
 
I solved #1 and got
Pythagorus says: .x² .= .(25t + 50)² + (18t)²

So we have: .x² .= .949t² + 2500t + 2500

Differentiate with respect to time: .2x(dx/dt) .= .1898t + 2500

. . and we have: .dx/dt .= .(1898t + 2500)/2x .[1]


At 7 PM (t = 7): .x² .= .949·7² + 2500·7 + 2500 .= .66,501
. . Hence: .x .= .√(66501)


Substitute into [1]: .dx/dt .= .(1898·7 + 2500)/(2√66501) .≈ .30.6 mph


so there is my attempt at work if you doubt me. Thanks for the help it got my started in the right direction! Please help some more!
 
anyone help with #2? I found what I think is the right equation but my answer keeps coming up wrong
 
I finished #2, I took the derivitive of the given equation and then solved for dV/dT. I got ((-410)^1.4(-11))/(75(1.4(410^.4)) . I am doing the work I promise I am just looking for tips!
 
so ... show us what you did (the calculus, not the numerical evaluation) and maybe someone will spot your mistake, if any.
 
I took the derivitive of PV^1.4=c which gave me P1.4V^.4(dV/dT)+410^1.4(dP/dT)=0 by the product rule and using implicit differentiation. After that I plugged in the given(s) and solved for dV/dT. The problem was easier than I thought. I solved #4 as well using the same technique. Now only #3 and #5 remain.

Thanks!
 
#3 sketch a picture ... you get a triangle

side 1 = radar station to the overhead point, 8 km

side 2 = overhead point to the plane's position at t = 1 min, call this distance S.

side 3 = radar station to plane's position at t = 1 min, call this distance D.

you want dD/dt at t = 1 min

you have dS/dt = 14 km/min

relationship between the three sides ... law of cosines

D<sup>2</sup> = S<sup>2</sup> + 8<sup>2</sup> - 2(8)(S)cos(140)

take the derivative w/r to time ... solve for dD/dt.
 
#5 ...

distance between any point (x,y) on the given curve and the origin is

\(\displaystyle \L D = \sqrt{(x-0)^2 + (y-0)^2} = (x^2 + y^2)^{\frac{1}{2}}\)

\(\displaystyle \L \frac{dD}{dt} = \frac{1}{2}(x^2 + y^2)^{-\frac{1}{2}}*(2x\frac{dx}{dt} + 2y\frac{dy}{dt})\)

clean up a bit ...

\(\displaystyle \L \frac{dD}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{(x^2 + y^2)^{\frac{1}{2}}}\)

now, you have dx/dt and x = 38(?) ... you can figure out y and dy/dt from your given function y = f(x). Finish up and solve for dD/dt.
 
Hello, rofl!

5. A particle is moving along the curve \(\displaystyle y\:=\:2\sqrt{5x\,+\,1}\).
As the particle passes through the point (3,8),
its x-coordinate increases at a rate of 4 units per second.
Find the rate of change of the distance from the particle to the origin at this instant.
Click to expand...

Let \(\displaystyle z\) = distance of the particle from the origin.

Then: \(\displaystyle \:z^2\:=\:x^2\,+\,y^2\;\;\)[1]

Differentiate with respect to time: \(\displaystyle \:2z\left(\frac{dx}{dt}\right) \:=\:2x\left(\frac{dx}{dt}\right)\,+\,2y\left(\frac{dy}{dt}\right)\)

. . and we have: \(\displaystyle \L\:\frac{dz}{dt} \:=\:\frac{1}{z}\left[x\left(\frac{dx}{dt}\right)\,+\,y\left(\frac{dy}{dt}\right)\right]\;\;\)[2]


Now we need all the quantities on the right side.

We know that: \(\displaystyle x\,=\,3,\:y\,=\,8,\:\frac{dx}{dt}\,=\,4\)
We need \(\displaystyle z\) and \(\displaystyle \frac{dy}{dt}\)

From [1], we have: \(\displaystyle \:z^2\:=\:3^2\,+\,8^2\;\;\Rightarrow\;\;z\,=\,\sqrt{73}\)

We are given: \(\displaystyle \:y\:=\:2(5x\,+\,1)^{\frac{1}{2}}\)
. . Hence: \(\displaystyle \,\frac{dy}{dt}\:=\:2\cdot\frac{1}{2}(5x\,+\,1)^{-\frac{1}{2}}\cdot5 \:=\:\frac{5}{\sqrt{5x\,+\,1}}\)
At (3, 8): \(\displaystyle \:\frac{dy}{dt}\:=\:\frac{5}{\sqrt{5\cdot3\,+\,1}}\:=\:\frac{5}{4}\)


Substitute all this into [2]: \(\displaystyle \L\:\frac{dz}{dt}\:=\:\frac{1}{\sqrt{73}}\left[3(4)\,+\,8\left(\frac{5}{4}\right)\right] \:=\;\frac{22}{\sqrt{73}}\)


Therefore, \(\displaystyle z\) is increasing at about \(\displaystyle 2.57\) units per second.

 
my mistake on writing the problem. It passes through (3,8) I understand everything you have done, but how do you find dY/dt?
 
Well with your help I solved the last one and the answer actually comes out to be
z=sqrt(8^2+3^2)=sqrt(73)

dZ/dt=2(3)(4)+2(8)(5)/2sqrt(73) which equals 52sqrt(73)/73
 
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