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The diagram shows the curve y=[tex]\frac{1}{16}[/tex] (3x - 1)², the point P (3, 4) lies on the curve and the tangent and normal at P(3,4) cuts the x- axis at Q and R respectively.
i. By using differentiation find the equation of tangent PQ and normal PR.
ii.By using vector product of two vectors QP and QR, find the area o the triangle PQR.

The diagram shows the curve ytexfrac116tex 3x 1 the point P 3 4 lies on the curve and the tangent and normal at P34 cuts the x axis at Q and R respectively i By class=

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LammettHash

i. Find dy/dx, which gives the slope of the tangent line to a point (x, y) on the curve.

[tex]y = \dfrac{(3x-1)^2}{16} \implies \dfrac{dy}{dx} = \dfrac{2(3x-1)\times3}{16}=\dfrac{9x-3}8[/tex]

At the point (3, 4), the slope of the tangent line is

[tex]\mathrm{slope}_{\rm tan} = \dfrac{dy}{dx}\bigg|_{x=3,y=4} = 3[/tex]

and the slope of the normal line is

[tex]\mathrm{slope}_{\rm nor} = -\dfrac1{\frac{dy}{dx}}\bigg|_{x=3,y=4} = -\dfrac13[/tex]

Both of these lines pass through the point (3, 4), so their equations are

[tex]y_{\rm tan} - 4 = 3 (x - 3) \implies \boxed{y_{\rm tan} = 3x - 5 ~~~~ (PQ)}[/tex]

[tex]y_{\rm nor} - 4 = -\dfrac13 (x - 3) \implies \boxed{y_{\rm nor} = -\dfrac x3 + 5~~~~ (PR)}[/tex]

ii. Find the x-intercepts of PQ and PR :

[tex]3x - 5 = 0 \implies x = \dfrac53 \implies \left(\dfrac53,0\right)[/tex]

[tex]-\dfrac x3 + 5 = 0 \implies x = 15 \implies (15, 0)[/tex]

Then the vector starting at Q and terminating at P is equivalent to the vector

[tex](3, 4) - \left(\dfrac53,0\right) = \left(\dfrac43,4\right)[/tex]

(though note that this vector starts at the origin)

and the vector starting at R and terminating at P is equivalent to

[tex](3,4) - (15,0) = (-12,4)[/tex]

Now, suppose we treat these two vectors as vectors in 3D space. They both lie in the plane, so the z-coordinates are 0.

Recall the cross product identity,

[tex]\|a\times b\| = \|a\| \|b\| \sin(\theta)[/tex]

where θ is the angle between vectors a and b. Here we know PQ and PR are perpendicular to one another so sin(θ) = sin(90°) = 1. The magnitude of their cross product corresponds to the area of the parallelogram spanned by them. Cut this area in half to get the area of ∆PQR.

The area is then

[tex]\dfrac12 \left\|\left(\dfrac43,4,0\right) \times (-12,4,0)\right\| \\\\ = \dfrac12 \left\|\left(\dfrac43,4,0\right)\right\| \|(-12,4,0)\| \\\\ = \dfrac12 \times \dfrac{4\sqrt{10}}3 \times 4\sqrt{10} \\\\ = \boxed{\dfrac{80}3}[/tex]

semsee45

Answer:

[tex]\textsf{i)} \quad \textsf{Tangent PQ}: \: y = 3x - 5[/tex]

     [tex]\textsf{Normal PR}: \: y=-\dfrac{1}{3}x+5[/tex]

[tex]\textsf{ii)} \quad \sf Area=\dfrac{80}{3} \:\: units^2[/tex]

Step-by-step explanation:

Given:

  • [tex]y=\dfrac{1}{16}(3x-1)^2[/tex]
  • P = (3, 4)

Part (i)

The derivative of the function f(x) gives the gradient of the tangent to the graph of y = f(x) at the point (x, y).

[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times \dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]

Differentiate the function of the curve by using the chain rule.

[tex]\textsf{Let }\:y=\dfrac{1}{16}u^2 \: \textsf{ where }\:u=3x-1[/tex]

Differentiate y and u separately:

[tex]\implies \dfrac{\text{d}y}{\text{d}u}=\dfrac{2}{16}u=\dfrac{1}{8}u[/tex]

[tex]\implies \dfrac{\text{d}u}{\text{d}x}=3[/tex]

Put everything back into the chain rule formula:

[tex]\implies \dfrac{\text{d}y}{\text{d}x}=\dfrac{1}{8}u \times 3=\dfrac{3}{8}u=\dfrac{3}{8}(3x-1)[/tex]

To find the gradient of the curve at point P, substitute x = 3 into the differentiated function:

[tex]\textsf{When }x=3, \quad \dfrac{\text{d}y}{\text{d}x}=\dfrac{3}{8}(3(3)-1)=3[/tex]

So the gradient of the tangent at P is 3.

The gradient of the normal is the negative reciprocal of the gradient of the tangent, so the gradient of the normal at P is -¹/₃ .

Substitute the found gradients and the coordinates of the P into the point-slope form of linear equation to find the equation of the tangent PQ and the normal PR:

Equation of the tangent PQ

[tex]\implies y-y_1=m(x-x_1)[/tex]

[tex]\implies y-4=3(x-3)[/tex]

[tex]\implies y=3x-5[/tex]

Equation of the normal PR

[tex]\implies y-y_1=m(x-x_1)[/tex]

[tex]\implies y-4=-\dfrac{1}{3}(x-3)[/tex]

[tex]\implies y=-\dfrac{1}{3}x+5[/tex]

Part (ii)

To find the coordinates of point Q, set the equation of the tangent PQ to zero and solve for x:

[tex]\implies 3x-5=0[/tex]

[tex]\implies x=\dfrac{5}{3}[/tex]

Therefore, Q = (⁵/₃, 0).

To find the coordinates of point R, set the equation of the normal PR to zero and solve for x:

[tex]\implies -\dfrac{1}{3}x+5=0[/tex]

[tex]\implies x=15[/tex]

Therefore, R = (15, 0).

Coordinates of vector QP:

[tex]\begin{aligned}\overrightarrow{\sf QP} & =\overrightarrow{\sf QO}+\overrightarrow{\sf OP}\\ & = \overrightarrow{\sf OP}-\overrightarrow{\sf OQ}\\ & = \sf (3,4)-\left(\dfrac{5}{3},0\right)\\ & = \sf \left(\dfrac{4}{3},4\right)\end{aligned}[/tex]

Coordinates of vector QR:

[tex]\begin{aligned}\overrightarrow{\sf QR} & =\overrightarrow{\sf QO}+\overrightarrow{\sf OR}\\ & = \overrightarrow{\sf OR}-\overrightarrow{\sf OQ}\\ & = \sf (15,0)-\left(\dfrac{5}{3},0\right)\\ & = \sf \left(\dfrac{40}{3},0\right)\end{aligned}[/tex]

Area of triangle PQR

[tex]\begin{aligned}\sf Area\:of\:\triangle PQR & = \sf \dfrac{1}{2} \left| \overrightarrow{\sf QP} \cdot \overrightarrow{\sf QR}\right|\\\\ & = \sf \dfrac{1}{2} \left| (x_{QP} \cdot y_{QR})-(x_{QR} \cdot y_{QP}) \right|\\\\ & = \sf \dfrac{1}{2}\left| \left(\dfrac{4}{3} \cdot 0\right)-\left(\dfrac{40}{3} \cdot 4 \right)\right|\\\\ & =\sf \dfrac{80}{3}\:\:units^2\end{aligned}[/tex]

Learn more about differentiation here:

https://brainly.com/question/27926244

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