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).

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

3 answers

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).
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!
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.