Let t=f(τ)be a differentiable real-valued function of τ, and let r(t)be a differentiable vector function with three components such that f(τ)is in the domain of rfor every value of τon some interval I. Prove that drdτ=drdtdtdτ. (This is Theorem 11.8.)

Short Answer

Expert verified

Ans:

drdτ=ddτ(r(t))=ddτx(t),y(t),z(t)=drdt·dtdτ

Step by step solution

01

Step 1. Given information: 

Part (a). t=f(τ)be a differentiable real-valued function of τ.

Part (b). r(t)is differentiable vector function with three components such that f(τ)is in the domain of rfor every value of τon some interval I.

02

Step 2. Proving:

Let r(t)=x(t),y(t),z(t)

The objective is to prove that drdτ=drdt,dtdτ

drdτ=ddτ(r(t))=ddτx(t),y(t),z(t)=ddτ(x(t)),ddτ(y(t)),ddτ(z(t))Derivative of a vector function.

=dxdt,dtdτ,dydt,dtdτ,dzdt,dtdτchain rule for scalar functions.

=dxdt,dydt,dzdtdtdτ Multiplication of a vector

function with a scalar function.

=drdt·dtdτ

Derivative of a vector function.

Thus drdτ=drdt·dtdτ

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