Two Calculus Questions

Question

1. the tangent line to the curve y=x²lnx is horizontal at x equal to :
2. If the position function is f(x)=3cos(2t) then the acceleration function is :

Answer 1

1. Slope horizontal line = 0. So, to get tangent points we have to find derivative:

y' = 2xlnx + x²(1/x) = 2xlnx + x = 0; x(2lnx +1) [= 0; x = 0 not in domain of y; 2lnx = -1; x = e^-1/2,

Tangent line is horizontal at x = e^-1/2

2.Velocity function v(t) = f'(t) = - 6sin(2t)

Accelaration function a(t) = v'(t) = - 12cos(2t)

Answer 2

Let's remember that a tangent line is another phrase for the word "slope" or derivative in the lingo of calculus. Since the tangent line is expected to be horizontal, then the slope is 0 (from the days of algebra/algebra 2).

But, since we know that slope is another way to talk about a derivative, then let's take the derivative and set it to zero...

y = x²ln(x)

dy/dx = y' = x² * 1/x + 2x*ln(x) ---> by way of the product rule

dy/dx = x + 2x*ln(x) = 0

Factor out x to see what we have...

x*(1 + 2ln(x)) = 0

Since we have two factors, either of them could be zero --> x = 0, or 1 + 2ln(x) = 0. Attempting to solve the second equation (i.e. root) gives...

2ln(x) = -1 --> ln(x) = -1/2 --> x = e^-1/2 = 1/√e

So we have two answers --> x = 0, or x = 1/√e which will yield a horizontal tangent line

The second problem uses the concept of position, velocity, and acceleration...

velocity = derivative of position with respect to time --> v = df/dt

acceleration = derivative of the velocity with respect to time --> a = dv/dt = d²f/dt²

f = 3cos(2t)

f' = v = 3 * 2 * (-sin(2t)) = -6sin(2t)

f" = a = -6 * 2 * (cos(2t)) = -12cos(2t) = 4 * f