right triangle 3,4,5 or 60,80,100
x= 60
y = 80
z = hypotenuse = 100
dx/dt = 80
dy/dt = 100
z^2 = x^2+y^2
2 z dz = 2 x dx + 2 y dy
z dz/dt = xdx/dt +y dy/dt
100 dz/dt =60*80+80*100
dz/dt = 6*8+80
dz/dt = 128 km/hr
x= 60
y = 80
z = hypotenuse = 100
dx/dt = 80
dy/dt = 100
z^2 = x^2+y^2
2 z dz = 2 x dx + 2 y dy
z dz/dt = xdx/dt +y dy/dt
100 dz/dt =60*80+80*100
dz/dt = 6*8+80
dz/dt = 128 km/hr
let the distance traveled by car Y by y km
their paths form a right-angled triangle.
Let the distance between them be D km
D^2 = x^2 + y^2
2D dD/dt = 2x dx/dt + 2y dy/dt
dD/dt = (x dx/dt + y dy/dt)/D
at the given case:
x = 60, y = 80 , dx/dt = 80, dy/dt = 100
D^2 = 60^2 + 80^2 = 10000
D = 100
dD/dt = (60(80) + 80(100))/100
= 128
The distance is changing at 128 km/h
check my arithmetic
Let's denote the distance between car X and car Y as D(t), where t represents time in hours. We want to find dD/dt, which is the rate at which D is changing with respect to time.
First, let's denote the position of car X as X(t), where X(t) is the distance travelled by car X at time t. From the given information, we have X(t) = 60 + 80t (since car X is 60 km east of point P and travelling east at a speed of 80 km/h).
Similarly, let's denote the position of car Y as Y(t), where Y(t) is the distance travelled by car Y at time t. From the given information, we have Y(t) = 80 + 100t (since car Y is 80 km north of point P and travelling north at a speed of 100 km/h).
Now, the distance between car X and car Y at time t is given by the Euclidean distance formula:
D(t) = sqrt((X(t))^2 + (Y(t))^2)
Let's take the derivative of both sides of the equation with respect to time t:
dD/dt = (1/2)*(X(t)^2 + Y(t)^2)^(-1/2) * (2*X(t)*X'(t) + 2*Y(t)*Y'(t))
Plugging in the values we have:
dD/dt = (1/2)*(60 + 80t)^2 + (80 + 100t)^2)^(-1/2) * (2*(60 + 80t)*80 + 2*(80 + 100t)*100)
Now, we can simplify this expression to get the rate at which the distance between car X and car Y is changing, dD/dt, in terms of time t.