Question

Calculate the tangent vector at time t=2 to the path in R³p(t)=(tcos(2πt),tsin(2πt),t²)

Find the directional derivative of the following function at the point (2,3) in the direction i−2j. f(x,y)=x² yln(x²+y² )

Answer 1

To find the tangent vector at t=2 for the path p(t), one must differentiate its components with respect to time and evaluate at t=2. For the directional derivative of a function, one computes the gradient and dots it with the normalized direction vector at the given point.

Step-by-step explanation:

To calculate the tangent vector at time t=2 for the path p(t)=(tcos(2πt),tsin(2πt),t²), we first need to find the derivative of p(t) with respect to time t. This derivative will give us the velocity vector, which is tangent to the path at any point in time. The velocity vector v(t) is given by:

• The derivative of the first component: d(tcos(2πt))/dt
• The derivative of the second component: d(tsin(2πt))/dt
• The derivative of the third component: d(t²)/dt

Let's compute these derivatives step by step:

1. For the first component: Use the product rule and chain rule to differentiate tcos(2πt), which gives us cos(2πt) - 2πtsin(2πt).
2. For the second component: Similarly, apply the product rule and chain rule to tsin(2πt), yielding sin(2πt) + 2πtcos(2πt).
3. For the third component: The derivative of t² is straightforward: 2t.

Substituting t=2 into these derivatives gives us the tangent vector at t=2.

The directional derivative of the function f(x,y)=x² yln(x²+y²) in the direction of the vector i - 2j at the point (2,3) is calculated by finding the gradient of f and then taking the dot product with the unit vector in the direction of i - 2j. Remember to normalize the direction vector before calculating the dot product.