Find the unit vectors that are parallel to the tangent line to the curve y=2sin⁡x at the point (π6,1)

Question

Stuart Rountree · Accepted Answer

(π6,1)&amp;nbsp;where&amp;nbsp;x=π6, the slope of the tangent line:f(x)=2sin⁡x⇒f′(x)=2cos⁡x⇒f′(π6)=2cos⁡(π6)=2⋅32=3The slope of the tangent line at this point is&amp;nbsp;m=3Since slope is the ratio of rise per run or&amp;nbsp;yx, then we can set an arbitrary vector with slope 4 by letting&amp;nbsp;x=1&amp;nbsp;and&amp;nbsp;y=3&amp;nbsp;or&amp;nbsp;v=[1,3]. This vector has &quot;slope&quot;&amp;nbsp;3&amp;nbsp;and is thus parallel to the tangent line.uv=v|v|=1312+(3)2=134[12,32]−uv=[−12,−32]Answer[12,32]&amp;nbsp;and&amp;nbsp;[−12,−32]

lalilulelo2k3eq · Accepted Answer

Step 1To answer this question, we need to know the slope of the tangent at (π6,1)Recall that: f′(a) is the slope of the tangent of y=f(x) at x=aWe know that the derivative of y=2sin⁡x is y′=2cos⁡xTherefore, the slope of the tangent at (π6,1) is2cos⁡(π6)=2⋅32=3Let us consider the vector v=vxi+vyjIf this vector is parallel to a line with slope 3 , we can writevyvx=±3vy=±3⋅vx⇒(Equation1)Since this is a unit vector, we can also write the following equationv{x}2+v{y}2=1Use ({Equation 1)(Equation 1) to replace the expression for vy in the above equationv{x}2+3v{x}2=14v{x}2=1Divide both sides by 4v{x}2=14 Take the square root of both sidesvx=±12 Substitute this in (Equation 1)), to getvy=±32AnswerThe two unit vectors parallel to the tangent at(π6,1)(π6,1)are±&amp;lt;12,32&amp;gt;

nick1337 · Accepted Answer

Step   {y=2sin⁡(x)}  {dydx=slope=2cos⁡(x)}x=π6  To find parrallel unit vectors to the line tangent ot point (π6,1) all that is really needed is the slope. At the point in question, slope is 3  Step 2  {y=3⋅x}⇒⇒(1,3)  Equation of a parrellel line. Point satisfying equation.  Step 3  |v―|=(Δx)2+(Δy)2=2  magnitude of the vector along our parrallel line, with end at the point (1,3)  Step 4   12⋅&amp;lt;1,3&amp;gt;  normalizing components of the vector at our chosen point, to find the unit vector. The positve and negative of this vector are parrallel.  NSK Answer  ∥u―∥=12i+32j,or−12i−32

Jazz Frenia · Accepted Answer

Step 1: Calculate the derivative of y with respect to x:dydx=2cosxStep 2: Evaluate the derivative at the given point:dydx|(π6,1)=2cos(π6)=3Step 3: The tangent line to the curve at the given point has a slope equal to dydx, which is 3.Step 4: Since the unit vector has a magnitude of 1, we normalize the tangent vector by dividing it by its magnitude:T=v‖v‖, where T is the unit vector parallel to the tangent line and v is the tangent vector.Step 5: The tangent vector is v=[13].Step 6: Calculate the magnitude of the tangent vector:‖v‖=12+32=2.Step 7: Divide the tangent vector by its magnitude to get the unit vector:T=v‖v‖=12[13]=[1232]Therefore, the unit vector parallel to the tangent line to the curve y=2sinx at the point (π6,1) is [1232].

xleb123 · Accepted Answer

To find the unit vectors that are parallel to the tangent line to the curve y=2sin(x) at the point (π6,1), we need to determine the derivative of the function and evaluate it at the given point. The derivative of y=2sin(x) with respect to x is:dydx=2cos(x)Next, we substitute the given x-coordinate, π6, into the derivative:dydx|x=π6=2cos(π6)=2·32=3This value represents the slope of the tangent line at the point (π6,1). To obtain the unit vector parallel to this tangent line, we normalize the derivative by dividing it by its magnitude. The magnitude of the derivative is 3, so the unit vector 𝐯 parallel to the tangent line is given by:𝐯=13(13)=(1333)=(131)Hence, the unit vector parallel to the tangent line to the curve y=2sin(x) at the point (π6,1) is (131).

Find the unit vectors that are parallel to the tangent

Answered question

Answer & Explanation

New Questions in Precalculus