At what points on the given curve x = 4t3, y = 3 + 8t − 10t2 does the tangent line have slope 1?

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;

and

Reasons:

The given parameters are;

x = 4·t³

y = 3 + 8·t - 10·t²

Reasons:

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

or t = -2

Which for gives;

When t = -2, we get;

x = 4×(-2)³ = -32

y = 3 + 8×(-2) - 10×(-2)² = -53

The other point is

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(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)

d

step-by-step explanation:

d

17.08

correct me if i am wrong.