Find the dy/dt for each pair of functions y=x^2-5x , x=t^2+3
Well, to find dy/dt, we need to use the chain rule. Let's start by finding dy/dx:
dy/dx = d/dx (x^2 - 5x)
= 2x - 5
Now, we know that x = t^2 + 3. Let's substitute this into dy/dx:
dy/dx = 2(t^2 + 3) - 5
= 2t^2 + 6 - 5
= 2t^2 + 1
Thus, dy/dt = dy/dx * dx/dt:
dy/dt = (2t^2 + 1) * 2t
= 4t^3 + 2t
And that's the derivative of y with respect to t, also known as dy/dt. Hope that helps, and remember that laughter is also a great way to differentiate yourself!
To find dy/dt, we need to use the chain rule and differentiate both y and x with respect to t, and then divide dy/dt by dx/dt.
Given that y = x^2 - 5x and x = t^2 + 3, we first find dy/dx:
dy/dx = d/dx(x^2 - 5x)
= 2x - 5
Next, we find dx/dt:
dx/dt = d/dt(t^2 + 3)
= 2t
Finally, we find dy/dt by dividing dy/dx by dx/dt:
dy/dt = (dy/dx)/(dx/dt)
= (2x - 5)/(2t)
Substitute x = t^2 + 3 into the equation to get dy/dt in terms of t:
dy/dt = (2(t^2 + 3) - 5)/(2t)
= (2t^2 + 4 - 5)/(2t)
= (2t^2 - 1)/(2t)
So, the expression for dy/dt is (2t^2 - 1)/(2t).
To find dy/dt for the pair of functions, we need to express y in terms of t instead of x, and then differentiate with respect to t.
Given that x = t^2 + 3, we can solve for t in terms of x:
x = t^2 + 3
Subtracting 3 from both sides:
x - 3 = t^2
Taking the square root of both sides:
√(x - 3) = t
Now we can substitute this expression for t into the equation y = x^2 - 5x:
y = (t^2 + 3)^2 - 5(t^2 + 3)
Expanding and simplifying:
y = t^4 + 6t^2 + 9 - 5t^2 - 15
Combining like terms:
y = t^4 + t^2 - 6
Finally, we can differentiate y with respect to t to find dy/dt:
dy/dt = d/dt(t^4 + t^2 - 6)
Using the power rule for differentiation, we have:
dy/dt = 4t^3 + 2t
Therefore, the dy/dt for the pair of functions y = x^2 - 5x, x = t^2 + 3 is dy/dt = 4t^3 + 2t.
from y = x^2 - 5x
dy/dx = 2x - 5
from x = t^2+3
dx/dt = 2t
dy/dt = (dy/dx)(dx/dt)
= (2x-5)(2t)