Answer:
(4/3, 4 5/9) and (-32, -53)
Step-by-step explanation:
When a curve is given as a set of parametric equations, as this one is, then the slope of the tangent line to the curve is
dy/dt
dy/dx = ------------
dx/dt
which here is
dy/dt 8 - 20t
dy/dx = ----------- = --------------
dx/dt 12t^2
If the slope at a certain point on this curve is 1, then we conclude that:
8 - 20t = 12t^2, or
12t^2 + 20t - 8 = 0, or
3t^2 + 5t - 2 = 0
We have to solve this equation for the parameter, t:
Here a = 3, b = 5 and c = -2, and so the discriminant is
b^2 - 4ac = 25 - 4(3)(-2), or 49, and the square root of that is 7.
Thus, the roots are:
-5 ± 7
t = --------- = 1/3 and t = -2
2(3)
Evaluate x and y twice, once each for each t value.
Case 1: t = 1/3
x = 4(1/3) and y = 3 + 8(1/3) - 10(1/3)^2, or
x = 4/3 and y = 3 + 8/3 - 10/9: (4/3, 4 5/9)
Case 2: t = -2
x = 4(-2)^3 and y = 3 + 8(-2) - 10(-2)^2, or y = 3 - 16 - 40, or y = -53.
This gives us the point (-32, -53)
The given quantities are presented in the form of parametric equations
with the t as the common variable.
The points on the curve where the tangent have a slope of 1 are;
[tex]\underline{\mathrm{ \left(\dfrac{4}{27}, \ 4\dfrac{5}{9} \right)}}[/tex]and [tex]\underline{(-32, \, -53)}[/tex]
Reasons:
The given parameters are;
x = 4·t³
y = 3 + 8·t - 10·t²
Reasons:
[tex]\dfrac{dx}{dt} = 12 \cdot t^2[/tex]
[tex]\dfrac{dy}{dt} = 8 - 20 \cdot t[/tex]
[tex]\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt}} = \dfrac{ 8 - 20 \cdot t}{ 12 \cdot t^2} = 1[/tex]
12·t² = 8 - 20·t
12·t² + 20·t - 8 = 0
Using a graphing calculator, and by factorization, we have;
4·(3·t² + 5·t - 2) = 0
3·t² + 5·t - 2 = 0
(3·t - 1)·(t + 2) = 0
[tex]t = \dfrac{1}{3}[/tex] or t = -2
Which for [tex]t = \dfrac{1}{3}[/tex] gives;
[tex]x = 4 \times \left(\dfrac{1}{3} \right)^3 = \dfrac{4}{27}[/tex]
[tex]y = 3 + 8 \times \dfrac{1}{3} - 10 \times \left(\dfrac{1}{3} \right)^2 = \dfrac{41}{9} = 4\dfrac{5}{9}[/tex]
[tex]\underline{\mathrm{ A \ point \ is \left(\dfrac{4}{27}, \ 4\dfrac{5}{9} \right)}}[/tex]
When t = -2, we get;
x = 4×(-2)³ = -32
y = 3 + 8×(-2) - 10×(-2)² = -53
The other point is [tex]\underline{(-32, \, -53)}[/tex]
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