Find an equation in x and y for the line tangent to the curve x(t) = 4t², y(t) = t⁴ at the point (10,16).

Question

A) y - 16 = 4(x - 10)
B) y = 4x - 10
C) y = 16x - 10
D) y = 4x + 10

Answer 1

To begin, let's find the derivative of both x(t) and y(t) with respect to t:

x'(t) = 8t
y'(t) = 4t³

Let's find the value of t at which the tangent line passes through the point (10, 16):

x(t) = 4t² gives us 4t² = 10, so t = √(10/4) = √2.

Now, let's find the slope of the tangent line at t = √2:

y'(t) = 4t³ gives us y'(√2) = 4(√2)³ = 8√2.

To find the equation of the tangent line, use the point-slope form:

y - 16 = 8√2(x - 10)

Simplify the equation:

y - 16 = 8√2x - 80√2
y = 8√2x - 80√2 + 16
y = 8√2x - 80√2 + 16 is a simplification of the equation.

Comparing with the answer choices, we can see that the equation in x and y for the line tangent to the curve is most closely represented by:

A) y - 16 = 4(x - 10)

So the correct answer is A) y - 16 = 4(x - 10).

Answer 2

first, you are wrong when you say
x(t) = 4t² gives us 4t² = 10, so t = √(10/4) = √2
in fact, t = √(5/2)
However, we also need y(t) = 16, so t^4 = 16, making t=2.
So, the point (10,16) is not on the curve at all!

Answer 3

Apologies for the mistake. You are correct.

Given the corrected information, let's find the derivative of both x(t) and y(t) with respect to t:

x'(t) = 8t
y'(t) = 4t³

Now, let's find the value of t at which the tangent line passes through the point (10, 16):

We are looking for the value of t for which x(t) = 10 and y(t) = 16:

From x(t) = 4t², we have 4t² = 10, so t² = 10/4 = 5/2.
Taking the square root, t = ± √(5/2).

However, we also need y(t) = 16:
From y(t) = t⁴, we have (± √(5/2))⁴ = 16.

Simplifying, (√(5/2))⁴ = 5²/2² = 25/4 ≠ 16.
(-√(5/2))⁴ = (-1)⁴ * (√(5/2))⁴ = (√(5/2))⁴ = 25/4 ≠ 16.

Therefore, there is no value of t for which the curve x(t) = 4t², y(t) = t⁴ passes through the point (10, 16).

So, this question is not solvable and none of the answer choices are correct.