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goldfiish [28.3K]
3 years ago
7

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the pa

rameter.x = 6 + ln(t), y = t2 + 6, (6, 7)
Mathematics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:2x-y=5

Step-by-step explanation:

Given

x=6+\ln t

y=t^{2}+6

\left ( a\right ) without eliminating parameter

\frac{\mathrm{d} x}{\mathrm{d} t}=\frac{1}{t}

\frac{\mathrm{d} y}{\mathrm{d} t}=2t

\frac{\mathrm{d} y}{\mathrm{d} x}=2t^2

at \left ( 6,7\right )

6=6+\ln\left ( t\right )

t=1

Equation of line is given by

2=\frac{y-7}{x-6}

2x-12=y-7

2x-y=5

\left ( b\right )by eliminating parameter

x-6=\ln \left ( t\right )

t=e^{x-6}

y=t^2 +6

y=e^{\left ( 2x-12\right )}+6

differentiating we get

\frac{\mathrm{d}y}{\mathrm{d} x}=2e^\left ( 2x-12\right )

at \left ( 6,7\right )

\frac{\mathrm{d}y}{\mathrm{d} x}=2

2\left ( x-6\right )=\left ( y-7\right )

2x-y=5

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The angle of elevation of 61° and 72° with the height of the tower being 553.3 m. gives Vic's distance from Dan as approximately 356 meters.

<h3>How can the distance between Vic and Dan be calculated?</h3>

Location of Vic relative to the tower = South

Vic's sight angle of elevation to the top of the tower = 61°

Dan's location with respect to the tower = West

Dan's angle of elevation in order to see the top of the tower = 72°

Height of the tower = 553.3m

tan( \theta) =  \frac{opposite}{adjacent}

tan( 61 ^{ \circ}) =  \frac{553.3}{ Vic' s\: distance \: to \: tower}

distance  =  \frac{553.3}{tan ( {61}^{ \circ} )}  = 306.7

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Similarly, we have;

Dan's distance  =  \frac{553.3}{tan ( {72}^{ \circ} )}  = 179.8

  • Dan's distance from the tower ≈ 179.8 m

Given that Vic and Dan are at right angles relative to the tower (Vic is on the south of the tower while Dan is at the west), by Pythagorean theorem, the distance between Vic and Dan <em>d </em>is found as follows;

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Therefore;

  • Vic is approximately 356 meters from Dan

Learn more about Pythagorean theorem here:

brainly.com/question/343682

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