What is dy/dx if x=5t^3 and y=3t^5?

Asked on by luna9219

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

y = 3t^5

x =5t^3.

Here both x and y are expressed in terms of t. This is called the parametric equation of the curve .

Therefore dy/dx  = (dy/dt)* dt/dx.  . Or

dy/dx = (dy/dt)/(dx/dt)

y = 3t^5. So dy/dt = (3t^5)' = 3*5*t^(5-1) = 15t^4.

x =5t^3 . So  dx/dt = (5t^3)' = 5*3t^(3-1) = 15t^2.

Therefore dy/dx = (dy/dt)/(dx/dt) = 15t^4/(15t^2) = t^2.

Therefore dy/dx = t^2 .


krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

The answer posted above in incomplete to the extent that the value of dy/dx is given in terms of t rather than x.

We can complete the answer by convering the value of dy/dt in terms of x.

It is given:

x = 5t^3


t = (x/5)^(1/3)

Substituting this value of t in the value of dy/dx we get:

dy/dx = t^2 = [(x/5)^(1/3)]^2

= (x/5)^(2/3)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We notice that x and y are expressions that contain the variable t.

We'll differentiate x and y with respect to t.

dy/dx = (dy/dt)/(dx/dt)

x=5t^3 and y=3t^5

dy/dt = (d/dt)(3t^5)

dy/dt = 15t^4 (1)

(dx/dt) = (d/dt)(5t^3)

(dx/dt) = 15t^2 (2)

Now, we'll insert (1) and (2) in the ratio:

dy/dx = 15t^4/15t^2

We'll simplify and we'll get:

dy/dx = t^2

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