At what points on the given curve x = 4t^3, y = 4 + 24t -...

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Transcript

00:01 In this question, we have been given the parametric equation of a curve, x is equals to 4d cube and y is equals to 4 plus 24 times of t minus 14 t squared.

00:13 We need to find the point on the curve where the tangent have slope 1.

00:18 So slope is given to be equals to 1.

00:22 So first of all, what i will do, i'll try to find what is dx over dt.

00:26 So it will be what? it will be just 12 times t square.

00:30 And from here, d .y over dt, it comes out to be 24 minus 28 times of t.

00:39 So hence, dy over d x, which can be written as d.

00:44 Y over dt, whole divided by d x over dt.

00:49 Okay.

00:50 So d y over d t is 24 minus 28 t divided by dx over dt is nothing but 12 times of t square.

01:01 So we have to find the corresponding values of x and y.

01:06 Okay.

01:08 So this is given to be equal to 1.

01:11 This is given.

01:13 So cross multiplied 24 minus 28 t is equals to 12t square.

01:19 So from this i can say 12 t squared plus 28 t minus 24 equals to 0.

01:27 This can be reduced as 3t square plus 70 minus 6 equals to 0.

01:35 And we can easily factorize the factor will be t plus 3 and 3t minus 2 equals to 0 implies t value is minus 3 and 2 over 3.

01:46 So corresponding to this we will get following value of x and y...

Question

At what points on the given curve x = 4t^3, y = 4 + 24t - 14t^2 does the tangent line have slope 1? (x, y) = (smaller x-value) (x, y) = (larger x-value)

Question

At what points on the given curve x = 4t^3, y = 4 + 24t - 14t^2 does the tangent line have slope 1? (x, y) = (smaller x-value) (x, y) = (larger x-value)

Please give Ace some feedback