The sides and diagonal of the rectangle above are strictly increasing with time. At the instant when x=4 and y=3, dx/dt=dz/dt and dy/dt=k(dz/dt). What is the value of k at that instant.
solved for z with pythagorean theorem and stuck at that step. z=5
He had it right up until he plugged in x, y, and z. He forgot to multiply each by 2. Answer is 1/3.
It's the same whether he multiply each by 2 or not. Cuz 2 can be cancelled.[Ex:2(x dx/dt+ y dy/dt)= 2 z dz/dt]
He forgot to plug in Z.
Using pythagorean theorem
z=(x^2+y^2)^(1/2)
z=(16+9)^(1/2)
z=5
Thus
4 + 3 k = 5
3k=1
k=1/3
answer is 1/3
To find the value of k at the instant when x=4 and y=3, we can analyze the given information about the rates of change.
First, let's define the variables:
- x represents the length of one side of the rectangle.
- y represents the width of the rectangle.
- z represents the diagonal of the rectangle.
- dx/dt represents the rate of change of the length with respect to time.
- dy/dt represents the rate of change of the width with respect to time.
- dz/dt represents the rate of change of the diagonal with respect to time.
- k is the constant in the equation dy/dt = k * (dz/dt).
Given that the sides and diagonal of the rectangle are strictly increasing with time, we can use the Pythagorean theorem to relate the three variables:
z^2 = x^2 + y^2
Differentiating both sides with respect to time t, we get:
2z * (dz/dt) = 2x * (dx/dt) + 2y * (dy/dt) ... [Differentiation rule for composite function]
Since dx/dt and dz/dt are equal at the given instant (dx/dt = dz/dt), we can substitute dy/dt with k * (dz/dt) in the equation:
2z * (dz/dt) = 2x * (dx/dt) + 2y * k * (dz/dt)
Now, let's substitute the values x=4, y=3, and z=5 into this equation:
2(5) * (dz/dt) = 2(4) * (dx/dt) + 2(3) * k * (dz/dt)
10 * (dz/dt) = 8 * (dx/dt) + 6k * (dz/dt)
Since dx/dt = dz/dt, we can simplify the equation:
10 * (dz/dt) = 8 * (dz/dt) + 6k * (dz/dt)
10 = 8 + 6k
Now, let's solve for k:
6k = 10 - 8
6k = 2
k = 2/6
k = 1/3
Therefore, at the instant when x=4 and y=3, the value of k is 1/3.
Well, well, well, looks like we have ourselves a mighty fine rectangle here! So, we've got x, y, and z going on, getting longer with time. Quite the growth spurt!
Alright, so at the instant when x=4 and y=3, we know that dx/dt=dz/dt. But hold your horses, my friend, we're not done yet!
Apparently, dy/dt=k(dz/dt). Hmm, seems like dy/dt is a sneaky one, playing tricks with us. But fear not! We will solve this puzzle.
Now, you mentioned that you've used the Pythagorean theorem and found that z=5. Bravo! That's one less thing to figure out.
So, let's put all this information in our clowny brain and get to the bottom of this. We have:
dx/dt = dz/dt
dy/dt = k(dz/dt)
z = 5
Now, let's plug in the values of x=4 and y=3 into these equations and see where they take us. Okay, here we go:
dx/dt = dz/dt ---> (4)' = (5)'
Well, guess what? When you take the derivative of a constant, it magically disappears! Poof! So, dx/dt and dz/dt are equal to zero. It's like they vanished into thin air.
Moving on to the next equation:
dy/dt = k(dz/dt) ---> (3)' = k(5)'
Hold on a sec! Since dz/dt is also equal to zero, k times zero is still... zero! That means dy/dt is also equal to zero. Things are getting pretty wild here!
So, my friend, at that instant when x=4 and y=3, the value of k is zero. Zero, zilch, nada. It's a comedy of zeroes!
Hope that puts a smile on your face, my dear friend!
Well I assume that z is the diagonal.
x^2+y^2 = z^2
2 x dx/dt + 2 y dy/dt =2 z dz/dt
4 dz/dt + 3 k dz/dt = dz/dt
4 + 3 k = 1
k = -1