A plane flies horizontally with a speed Of 600km/h at an altitude of 10km and pasess directly over th town of quinton .find the rate at which the distance from the plane to quinton is increasing when its 20km away from quinton ?
let the horizontal distance from Quinton be x km
let the distance between plane and Quinto by y km
x^2 + 10^2 = y^2
2x dx/dt = 2y dy/dt
dy/dt = (x dx/dt)/y
when x=20
y^2 = 400 + 100 = 500
y = √500 = 10√5
dx/dt = 600 km/h
dy/dt = 20(600)/10√5 = appr. 536.7 km/h
yes, you can see in my first line of the solution that I used the Pythagorean theorem.
I then differentiate with respect to time, since we were talking about "rates" in the question and I saw a rate given as 600 km/h
Thank you !
how did you get dy/dt to be (x dx/dt)/y
Is it possible to use the trig laws such as using the tan to find the angle and then cosine to figure the hypotenuse? Because for the following question… it is not possible to use related rates. Instead you get the right answer of 2.35 m/s by using the trig laws…
A waterskier skis over the ramp shown in the figure at a speed of 12 m/s. How fast is she rising as she leaves the ramp? The horizontal length of the ramp given was 5 m while it was 1 m tall.
To find the rate at which the distance from the plane to Quinton is increasing when it's 20 km away, we can use the concept of similar triangles. Let's assume that at a given time, the distance from the plane to Quinton is represented by x km.
Let's visualize the scenario: The plane is flying horizontally above Quinton, forming a right-angled triangle with the altitude of 10 km. The distance from the plane to Quinton would be the hypotenuse of this triangle.
We have a right-angled triangle with sides 10 km and x km, and we want to find the rate at which x is increasing when x is 20 km.
Using the Pythagorean theorem, we can establish the relationship between the sides of the triangle:
x^2 + 10^2 = (20)^2
Simplifying the equation, we have:
x^2 + 100 = 400
x^2 = 300
To find the value of x, we take the square root of both sides:
x = √300
Now, we differentiate both sides of the equation implicitly with respect to time (t) to find the rate at which x is increasing:
d(x)/dt = d(√300)/dt
To differentiate √300, we can use the power rule of differentiation. Since 300 is a constant, its derivative is zero. Therefore, the differentiation simplifies to:
d(x)/dt = 0 /dt
d(x)/dt = 0
So, when the distance from the plane to Quinton is 20 km, the rate at which this distance is increasing is 0 km/h.
In conclusion, when the plane is 20 km away from Quinton, the rate at which the distance from the plane to Quinton is increasing is 0 km/h.