Derivatives of Vector-Valued Functions

Learning Outcomes

  • Write an expression for the derivative of a vector-valued function

Now that we have seen what a vector-valued function is and how to take its limit, the next step is to learn how to differentiate a vector-valued function. The definition of the derivative of a vector-valued function is nearly identical to the definition of a real-valued function of one variable. However, because the range of a vector-valued function consists of vectors, the same is true for the range of the derivative of a vector-valued function.

Definition


A derivative of a vector-valued function r(t)r(t) is

r(t)=limΔt0r(t+Δt)r(t)Δt,r(t)=limΔt0r(t+Δt)r(t)Δt,

 

provided the limit exists. If r(t)r(t) exists, then rr is differentiable at tt. If r(t)r(t) exists for all tt in an open interval (a, b)(a, b), then rr is differentiable over the interval (a, b)(a, b). For the function to be differentiable over the closed interval [a, b][a, b], the following two limits must exist as well:

r(a)=limΔt0+r(a+Δt)r(a)Δtr(a)=limΔt0+r(a+Δt)r(a)Δt and r(b)=limΔt0r(b+Δt)r(b)Δtr(b)=limΔt0r(b+Δt)r(b)Δt

 

Many of the rules for calculating derivatives of real-valued functions can be applied to calculating the derivatives of vector-valued functions as well. Recall that the derivative of a real-valued function can be interpreted as the slope of a tangent line or the instantaneous rate of change of the function. The derivative of a vector-valued function can be understood to be an instantaneous rate of change as well; for example, when the function represents the position of an object at a given point in time, the derivative represents its velocity at that same point in time.

We now demonstrate taking the derivative of a vector-valued function.

Example: Finding the derivative of a vector-valued function

Use the definition to calculate the derivative of the function

r(t)=(3t+4)i+(t24t+3)jr(t)=(3t+4)i+(t24t+3)j

TRY IT

Use the definition to calculate the derivative of the function r(t)=(2t2+3)i+(5t6)j.r(t)=(2t2+3)i+(5t6)j.

Notice that in the calculations in the example above, we could also obtain the answer by first calculating the derivative of each component function, then putting these derivatives back into the vector-valued function. This is always true for calculating the derivative of a vector-valued function, whether it is in two or three dimensions. We state this in the following theorem. The proof of this theorem follows directly from the definitions of the limit of a vector-valued function and the derivative of a vector-valued function.

Differentiation of Vector-Valued functions theorem


Let ff, gg, and hh be differentiable functions of tt.

  • If r(t)=f(t)i+g(t)jr(t)=f(t)i+g(t)j, then r(t)=f(t)i+g(t)jr(t)=f(t)i+g(t)j.
  • If r(t)=f(t)i+g(t)j+h(t)kr(t)=f(t)i+g(t)j+h(t)k, then r(t)=f(t)i+g(t)j+h(t)kr(t)=f(t)i+g(t)j+h(t)k.

Since we will be using derivatives of common single-variable functions throughout this course, we briefly recall the derivatives of common functions, along with frequently-encountered derivative rules below.

Recall: Derivatives of common functions

  • ddx(xn)=nxn1ddx(xn)=nxn1  (Power Rule)
  • ddx(sinx)=cosxddx(sinx)=cosx
  • ddx(cosx)=sinxddx(cosx)=sinx
  • ddx(secx)=secxtanxddx(secx)=secxtanx
  • ddx(cscx)=cscxcotxddx(cscx)=cscxcotx
  • ddx(tanx)=sec2xddx(tanx)=sec2x
  • ddx(cotx)=csc2xddx(cotx)=csc2x
  • ddx(ex)=exddx(ex)=ex
  • ddx(lnx)=1xddx(lnx)=1x
  • ddx(arctanx)=11+x2ddx(arctanx)=11+x2

Recall: Derivative Rules

Constant Multiple Rule: ddx(cf(x))=cf(x)ddx(cf(x))=cf(x)

Sum and Difference Rule: ddxf(x)±g(x))=f(x)±g(x)ddxf(x)±g(x))=f(x)±g(x)

Product Rule: ddx(f(x)g(x))=g(x)f(x)+f(x)g(x)ddx(f(x)g(x))=g(x)f(x)+f(x)g(x)

Quotient Rule: ddx(f(x)g(x))=g(x)f(x)f(x)g(x)[g(x)]2ddx(f(x)g(x))=g(x)f(x)f(x)g(x)[g(x)]2

Chain Rule: ddxf(g(x)=f(g(x))g(x)ddxf(g(x)=f(g(x))g(x)

Example: calculating the derivative of vector-valued functions

Use Differentiation of Vector-Valued Functions to calculate the derivative of each of the following functions.

  1. r(t)=(6t+8)i+(4t2+2t3)jr(t)=(6t+8)i+(4t2+2t3)j
  2. r(t)=3costi+4sintjr(t)=3costi+4sintj
  3. r(t)=etsinti+etcostje2tkr(t)=etsinti+etcostje2tk

TRY IT

Calculate the derivative of the function

r(t)=(tlnt)i+(5et)j+(costsint)kr(t)=(tlnt)i+(5et)j+(costsint)k

Watch the following video to see the worked solution to the above Try It

We can extend to vector-valued functions the properties of the derivative. In particular, the constant multiple rule, the sum and difference rules, the product rule, and the chain rule all extend to vector-valued functions. However, in the case of the product rule, there are actually three extensions: (1) for a real-valued function multiplied by a vector-valued function, (2) for the dot product of two vector-valued functions, and (3) for the cross product of two vector-valued functions.

Differentiation of Vector-Valued functions theorem


Let rr and uu be differentiable vector-valued functions of tt, let ff be a differentiable real-valued function of

1.ddt[cr(t)]=cr(t)Scalar multiple2.ddt[r(t)±u(t)]=r(t)±u(t)Sum and difference3.ddt[f(t)u(t)]=f(t)u(t)+f(t)u(t)Scalar product4.ddt[r(t)u(t)]=r(t)u(t)+r(t)u(t)Dot product5.ddt[r(t)×u(t)]=r(t)×u(t)+r(t)×u(t)Cross product6.ddt[r(f(t))]=r(f(t))f(t)Chain rule7.r(t)r(t)=c, then r(t)r(t)=0.1.ddt[cr(t)]=cr(t)Scalar multiple2.ddt[r(t)±u(t)]=r(t)±u(t)Sum and difference3.ddt[f(t)u(t)]=f(t)u(t)+f(t)u(t)Scalar product4.ddt[r(t)u(t)]=r(t)u(t)+r(t)u(t)Dot product5.ddt[r(t)×u(t)]=r(t)×u(t)+r(t)×u(t)Cross product6.ddt[r(f(t))]=r(f(t))f(t)Chain rule7.r(t)r(t)=c, then r(t)r(t)=0.

Proof

The proofs of the first two properties follow directly from the definition of the derivative of a vector-valued function. The third property can be derived from the first two properties, along with the product rule from the Introduction to Derivatives. Let u(t)=g(t)i+h(t)ju(t)=g(t)i+h(t)j. Then

ddt[f(t)u(t)]=ddt[f(t)(g(t)i+h(t)j)]=ddt[f(t)g(t)i+f(t)h(t)j]=ddt[f(t)g(t)]i+ddt[f(t)h(t)]j=(f(t)g(t)+f(t)g(t))i+(f(t)h(t)+f(t)h(t))j=f(t)u(t)+f(t)u(t)ddt[f(t)u(t)]=ddt[f(t)(g(t)i+h(t)j)]=ddt[f(t)g(t)i+f(t)h(t)j]=ddt[f(t)g(t)]i+ddt[f(t)h(t)]j=(f(t)g(t)+f(t)g(t))i+(f(t)h(t)+f(t)h(t))j=f(t)u(t)+f(t)u(t).

 

To prove property 4. Let r(t)=f1(t)i+g1(t)jr(t)=f1(t)i+g1(t)j and u(t)=f2(t)i+g2(t)ju(t)=f2(t)i+g2(t)j. Then

ddt[r(t)u(t)]=ddt[f1(t)f2(t)+g1(t)g2(t)]=f1(t)f2(t)+f1(t)f2(t)+g1(t)g2(t)+g1(t)g2(t)=f1(t)f2(t)+g1(t)g2(t)+f1(t)f2(t)+g1(t)g2(t)=(f1i+g1j)(f2i+g2j)+(f1i+g1j)(f2i+g2j)=r(t)u(t)+r(t)u(t).

 

The proof of property 5 is similar to that of property 4. Property 6 can be proved using the chain rule. Last, property 7 follows from property 4:

ddt[r(t)r(t)]=ddt[c]r(t)r(t)+r(t)r(t)=02r(t)r(t)=0r(t)r(t)=0.

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Now for some examples using these properties.

Example: Using the properties of derivatives of vector-valued functions

Given the vector-valued functions

r(t)=(6t+8)i+(4t2+2t3)j+5tk

and

u(t)=(t23)i+(2t+4)j+(t33t)k,

calculate each of the following derivatives using the properties of the derivative of vector-valued functions.

  1. ddt[r(t)u(t)]
  2. ddt[u(t)×u(t)]

TRY IT

Given the vector-valued functions r(t)=costi+sintje2tk and u(t)=ti+sintj+costk, calculate ddt[r(t)r(t)] and ddt[u(t)×r(t)].

Watch the following video to see the worked solution to the above Try It