South Central High Students' Council is traveling to Redding for the Students' Council National Conference. From the travel budg
et allowed for the trip, they have two options. They can leave by express train tonight, or they can save one hour by leaving tomorrow morning and flying which travels 40 km/h faster than the train. If the distance between South Central High and Redding is 480 km, determine how long it will take to travel by flying, rounded to the nearest hour.
2 answers:
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a. The equation of the tangent at (1,3) is y = -x/3 + 10/3
b. The equation of the normal at (1,3) is y = 3x
c. Find the graph in the attachment
<h3>a. How to find the equation of the tangent at (1, 3)?</h3>Since x² + y² = 10, we differentiate the equation with respect to x to find dy/dx which the the equation of the tangent.
So, x² + y² = 10
d(x² + y²)/dx = d10/dx
dx²/dx + dy²/dx = 0
2x + 2ydy/dx = 0
2ydy/dx = -2x
dy/dx = -2x/2y
dy/dx = -x/y
At (1,3), dy/dx = -1/3
Using the equation of a straight line in slope-point form, we have
m = (y - y₁)/(x - x₁) where
- m = gradient of the tangent = dy/dx at (1,3) = -1/3 and
- (x₁, y₁) = (1,3)
So, m = (y - y₁)/(x - x₁)
-1/3 = (y - 3)/(x - 1)
-(x - 1) = 3(y - 3)
-x + 1 = 3y - 9
3y = -x + 1 + 9
3y = -x + 10
3y + x = 10
y = -x/3 + 10/3
So, the equation of the tangent at (1,3) is y = -x/3 + 10/3
<h3>b. The equation of the normal at the point (1, 3)</h3>Since the tangent and normal line are perpendicular at the point, for two perpendicular line,
mm' = -1 where
- m = gradient of tangent = -1/3 and
- m' = gradient of normal
So, m' = -1/m
= -1/(-1/3)
= 3
Using the equation of a straight line in slope-point form, we have
m' = (y - y₁)/(x - x₁) where
- m' = gradient of normal at (1, 3) and (
- x₁, y₁) = (1,3)
So, m = (y - y₁)/(x - x₁)
3 = (y - 3)/(x - 1)
3(x - 1) = (y - 3)
3x - 3 = y - 3
y = 3x - 3 + 3
y = 3x + 0
y = 3x
So, the equation of the normal at (1,3) is y = 3x
c. Find the graph in the attachment
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