Derivatives of vector-valued functions Differentiate the following function r(t)=\tan ti+\sec tj+\cos^{2}tk

crapthach24

crapthach24

Answered question

2021-12-04

Derivatives of vector-valued functions
Differentiate the following function
r(t)=tanti+sectj+cos2tk

Answer & Explanation

Lorraine David

Lorraine David

Beginner2021-12-05Added 13 answers

Step 1
The given function is:
r(t)=tan(t)i+sec(t)j+cos2(t)k
We have to differentiate the given function.
The given function can be written as:
1) r(t)=tan(t)i+sec(t)j+(cos(t))2k
Step 2
Now differentiate the both sides of the equation (1) with respect to 't',
ddt(r(t))=ddt(tan(t)i+sec(t)j+(cos(t))2k)
r(t)=ddt(tan(t))i+ddt(sec(t))j+ddt(cos(t))2k
r(t)=sec2(t)i+sec2(t)i+sec(t)tan(t)j+2cos(t)×ddt(cos(t))k
r(t)=sec2(t)i+sec(t)tan(t)j+2cos(t)×(sin(t))k
r(t)=sec2(t)i+sec(t)tan(t)j2sin(t)cos(t)k
r(t)=sec2(t)i+sec(t)tan(t)jsin(2t)k
(sin2θ=2sinθcosθ)
therefore the value of the function after differentiation is:

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