Let 𝐫(t)=⟨3 cost, 5 sint, 4 cost⟩. Show that 𝐫(t) is constant and conclude, using Example 7, tha

Let 𝐫(t)=⟨3 cost, 5 sint, 4 cost⟩. Show that 𝐫(t) is...

Key Concepts

Vector-Valued Functions

A vector-valued function assigns a vector to each value of its parameter, frequently representing the position of a point in space as the parameter changes. This concept is foundational in describing curves and motions in multivariable calculus, where one can analyze not only the position but also the velocity and acceleration of a moving object via derivatives.

Norm (Magnitude) and Constant Norm

The norm of a vector, often denoted ||r(t)||, measures its length. When the norm is constant as a function of the parameter, it implies that the vector function's output always lies on the surface of a sphere centered at the origin with that constant radius. This invariant length property is critical because it leads to the conclusion that the derivative of the vector function is orthogonal to the function itself, reflecting that the point is moving tangentially along the sphere.

Derivative of a Vector Function

Taking the derivative of a vector-valued function with respect to its parameter yields another vector that represents the rate of change or velocity of the original function. This derivative captures both the speed and the direction of motion along the curve, and its computation typically involves differentiating each component of the vector function individually.

Orthogonality and Dot Product

Orthogonality in the context of vectors means that two vectors are perpendicular to each other. This condition is mathematically expressed by their dot product being zero. In problems where the vector function has a constant norm, differentiating the squared norm leads to an expression that shows the dot product of the vector and its derivative is zero, directly confirming that these vectors are orthogonal.

Transcript

00:01 So we have a vector where root function r of t is equal to 3 cos t comma 5 sine t comma 4 cos t and we have to find the norm of the vector r of t so this is equal to square root of 3 t whole square as 5 sine t whole square plus 4 cost t whole square this is equal to square root of nine cos square t plus 25 sine square t plus 16 cost square t so this can be written as nine cost square t plus nine sine square t plus 16 sine square t plus 16 cos square t this is equal to square root of 9 of cos square t plus sine square t plus 16 of sine square t plus coss square t this is equal to root over of 9 plus 16 is equal to root 25 is equal to 5 this is using fact that sine square t plus cos square t is equal to 1 the sense norm of r of t is equal to 5 is constant, we can find the derivative of r of t, derivative of r of t whole square, norm of the whole square.

02:32 So using product rule this will be equal to d by d t of r of t, ready, dot product with r of t.

03:02 Using the theorem that norm of vector a is equal to a dot a now using product rule this will be equal to r of t dot r d dot r d thus r d dot r of t so this will be equal to r of t s r of t dot r d is equal two times of r of t dot r d now we have used the property that a dot b is equal to b dot a so since this r dot t is con form of r dot is constant so d by d t of r of t of t of t whole square is equal to zero this is equal to to 2 times of r of r of t r d this implies r of t dot r dash t is equal to 0 and since 1 1 1 2 vectors is 0 so r t and r dash t are orthogonal now we can find r d separately and this is valid so r test d will be equal to d by d t 3 cost t comma 5 sign t comma 4 cost t this is equal to d by d d t of 3 cost t t by d t of five sign t come a t by d t of four cost t this is equal to minus three sign p five cost t come minus four sign t so this is r d r of t now r of t dot r d is t is equal to three cost t 5 sine t 4 cost t dot dot product with minus 3 sine t 5 cost t comma minus 4 sign t the other dot product will give us minus 9 cost t sine t that's 25 sign t and 16 cost t, 6, c...

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