Suppose that x and y are differentiable functions of t and...

Suppose that $x$ and $y$ are differentiable functions of $t$ and that $144 = x^2 + y^2$, then $\frac{dy}{dt} = \boxed{\phantom{000}}$ Use either $\frac{dx}{dt}$ or $x'(t)$ in your answer. If $\frac{dx}{dt} = -7$ when $x = 5$ and $y > 0$, then $\frac{dy}{dt}|_{x=5} = \boxed{\phantom{000}}$

Transcript

00:01 In this question, we are given an equation x squared plus y squared equals to 400, where x and y are both functions of time.

00:10 X is a function of t and y is also a function of t.

00:15 They are not independent variables.

00:18 We are asked to calculate dy over dt given some conditions and dx over dt also given some conditions.

00:25 But first, what we are going to do first is we will differentiate the equation.

00:30 We will calculate d over d t of x squared plus y squared and we will also differentiate the right hand side.

00:46 So we differentiated both sides of the equation and on the left hand side we are going to get 2x.

00:54 But since x is a function of time, by the chain rule, we need to multiply this by dx over d t plus 2y.

01:04 And since y is also a function of time, depends on time, we need to multiply this by dx over d t.

01:09 D t and the derivative of of 400 is 0 because 400 is a constant and we can cancel everything by 2 we can divide everything by 2 and we are going to get that x times d x over d t plus y x x x x x x x equals to 0 now we can move on to answering the questions in the first question we are asked to find d y over d t so we need to solve the equation for d y over d t from this equation here d .y or dt equals to negative y times dy or dt.

02:09 So, all right.

02:11 Y times d .y or dt from this equation equals to negative x times dx over dt.

02:21 And therefore, d .y or dt after dividing by y equals to negative x over y multiplied by dx or dt...

Question

Suppose that $x$ and $y$ are differentiable functions of $t$ and that $144 = x^2 + y^2$, then $\frac{dy}{dt} = \boxed{\phantom{000}}$ Use either $\frac{dx}{dt}$ or $x'(t)$ in your answer. If $\frac{dx}{dt} = -7$ when $x = 5$ and $y > 0$, then $\frac{dy}{dt}|_{x=5} = \boxed{\phantom{000}}$

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