Implicitly Defined Parametrizations

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Implicitly Defined Parametrizations

Assuming that the equations in Exercises 15−20 define x and y implicitly as differentiable functions x=f(t), y=g(t), find the slope of the curve x=f(t), y=g(t) at the given value of t.

x sin t + 2x = t, t sin t − 2t = y, t = π

Accepted Answer

Welcome back everyone. Assuming that the following equations define x and y implicitly as differentiable functions of t, find the slope of the curve at t equals pi. X cosine c plus 3 x equals t square and t cosine c + 4 t equals y. A pi divided by 3, b 3 divided by pi, C 3 pi and d 0. So for this problem, let's remember how we find the derivative of y with respect to x given. Equations In terms of y of T and XT. So basically, DY divided by DX is going to be equal to DY divided by DC divided by. DX divided by DC. So what we want to do is identify the derivative of Y with respect to t, first of all, right? And we can perform differentiation of the equation y equals c cosine c plus 4 c. So DY divided by DC. Is going to be equal to the derivative oft cosine c plus 4 t. Let's use the product rule for the first term, so we're going to get the derivative cosine C plus. T multiplied by cosinet derivative plus 4 t derivative for the next term. We get cosine C. The derivative of cosine is negative sign, so we get minus c sinc. + 4. Well done, and now let's find the derivative of x with respect to t. Essentially what we can do is differentiate implicitly, right? So we're going to take the derivative of the left-hand side and the right-hand side. On the left-hand side, we have the derivative of X cosine C. Plus 3 x derivative. Equals The derivative of t 2. Well done and now let's find each draw of a dove. We get the derivative of x multiplied by cosinec plus x multiplied by cosinec derivative, plus 3 x derivative equals 2 t. And now we can find the derivative of X, right? So we're going to factor out x. We get x and Cs cosine C. + 3. Plus X cosinec derivative is basically. Negative Xin C. And this is equal to 2 t. Now let's add x sin t to both sides. So that x cosine c + 3 equals. 2 T. Plus X C. And now X, which is DX divided by DT, is going to be equal to the right hand side. 2 T. Plus X Sign. See Divided by The factor cosine C + 3. And now from here. Let's find the derivative Dy divided by dx when C is equal to pi, right? So what we want to do is identify DY divided by DC. When t is equal to pi. That would be cosine of pi minus. By Sign of pi. Plus 4. And this is going to be equal to now cosine of pi is -1. Sin of pi is 0, so we can cross out the middle term, and we basically get -1 + 4, which is 3. Now for DX divided by DT. Let's evaluate S T is equal to pi, that would be. 2 pai Plus For X, let's just leave it as X for now. Notice that it's multiplied by sine of pi and sine of pi is 0, so our x term simply vanishes and we're dividing by. Co-sign off. 1 + 3. This is equal to 2 pi divided by cosine of pi is -1 + 3 gives us 2, so we get pi. Well done. And now let's find the derivative. So Dy divided by dt when t is equal to pi is the ratio between 3 and pi. This means that the correct answer is option B. Thank you for watching.

Implicitly Defined ParametrizationsAssuming that the equations in Exercises 15−20 define x and y implicitly as differentiable functions x=f(t), y=g(t), find the slope of the curve x=f(t), y=g(t) at the given value of t.x sin t + 2x = t, t sin t − 2t = y, t = π

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Implicitly Defined Parametrizations

Assuming that the equations in Exercises 15−20 define x and y implicitly as differentiable functions x=f(t), y=g(t), find the slope of the curve x=f(t), y=g(t) at the given value of t.

x sin t + 2x = t, t sin t − 2t = y, t = π