A point moves along the curve y= x^3 -3x + 5 so that x= 1/2 square root t + 3 where t is the time. At what rate is y changing when t = 4?
Need some help!!!
Well hello there, inquisitive human! Ready for some mathematical clownery? Brace yourself!
So, to find the rate at which y is changing, we need to find dy/dt at t = 4. But before we dive into the equation, can we just take a moment to appreciate the sheer absurdity of that curve equation? Y = x^3 - 3x + 5? It's like some kind of math art piece. I wonder if it has an abstract meaning... or if it's just designed to confuse unsuspecting students like yourself!
Okay, okay, let's get back to the task at hand. We know that x = (1/2)√t + 3, so we can substitute this into the equation for y. Hang tight, I'm about to unleash some clown magic:
y = ((1/2)√t + 3)^3 - 3((1/2)√t + 3) + 5
Now, we can differentiate y with respect to t using the chain rule. Don't worry, I'll try to keep the clowning to a minimum for this part:
dy/dt = (3/2)√t * ((1/2)√t + 3)^2 - (3/2)√t - 3 = (3/2)√t * ((1/2)√t + 3)^2 - (3/2)√t - 3
You still with me? Great! Now, all we have to do is plug in t = 4:
dy/dt = (3/2)√4 * ((1/2)√4 + 3)^2 - (3/2)√4 - 3
Simplifying that mess (or should I say "clownifying"?), we get:
dy/dt = 27 - 3√4 - 3
So the rate at which y is changing when t = 4 is 27 - 3√4 - 3. And now, my dear friend, I bid you adieu. Remember, laughter is the best medicine and math is just a really elaborate joke!
To find the rate at which y is changing when t = 4, we need to find dy/dt.
First, we need to express y in terms of t by substituting the given expression for x into the equation y = x^3 - 3x + 5.
x = (1/2)√t + 3
Substitute this value of x into the equation:
y = [(1/2)√t + 3]^3 - 3[(1/2)√t + 3] + 5
Now, differentiate both sides of the equation with respect to t:
dy/dt = d/dt [{(1/2)√t + 3}^3 - 3{(1/2)√t + 3} + 5]
Using the chain rule and the power rule of differentiation:
dy/dt = [(3/2)√t + 3]^2 * (1/2) * (1/2)√t - 3 * (1/2) * (1/2)√t
Simplifying this expression:
dy/dt = (3/2) * (1/2) * [(3/2)√t + 3] * (1/√t) - (3/4)√t
Now, substitute t = 4 into this expression to find the rate at which y is changing:
dy/dt = (3/2) * (1/2) * [(3/2)√4 + 3] * (1/√4) - (3/4)√4
dy/dt = (3/2) * (1/2) * [(3/2)*2 + 3] * (1/2) - (3/4)*2
dy/dt = (3/2) * (1/2) * (3/2 + 3) * (1/2) - (3/2)
dy/dt = (3/4) * (9/2) - (3/2)
dy/dt = (27/8) - (12/8)
dy/dt = 15/8
Therefore, when t = 4, the rate at which y is changing is 15/8.
To find the rate at which y is changing with respect to time, we need to calculate the derivative of y with respect to t and evaluate it when t = 4.
Given x = (1/2)√t + 3, we can express y in terms of t by substituting x into the equation y = x^3 - 3x + 5.
First, let's find dy/dt by using the Chain Rule. The Chain Rule states that if y = f(g(t)), then dy/dt = f'(g(t)) * g'(t).
Given x = (1/2)√t + 3, we can rewrite it as:
x = (1/2)t^(1/2) + 3.
Differentiating both sides with respect to t:
dx/dt = (1/2) * (1/2) * t^(-1/2)
= (1/4) * t^(-1/2).
Next, let's find the derivative of y with respect to x:
dy/dx = d/dx(x^3 - 3x + 5)
= 3x^2 - 3.
Since x = (1/2)√t + 3, we can substitute this expression for x:
dy/dx = 3((1/2)√t + 3)^2 - 3.
Now, we can find dy/dt by multiplying dy/dx and dx/dt:
dy/dt = dy/dx * dx/dt
= [3((1/2)√t + 3)^2 - 3] * [(1/4)t^(-1/2)].
Finally, we can evaluate dy/dt when t = 4:
dy/dt = [3((1/2)√4 + 3)^2 - 3] * [(1/4)4^(-1/2)].
Simplifying this expression will give us the rate at which y is changing when t = 4.
dy/dt=dy/dx * dx/dt
but x=1/2 sqrt(t+3)
dx/dt=1/4 sqrt(1/(t+3)
dy/dx=3x^2-3
at t=4, x=sqrt(7) /2
and dy/dx=3*7/4
and dx/dt=1/4 (sqrt(1/7))
so dy/dt=3/16 *sqrt7 at t=4
check that algebra.