PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION
277
34.18
Find dy/dx and d
2 y/dx
2 for x = t* + t, y = i + t+\.
dx/dt = 3t
2 + l, dy/dt = 7t
6 + l. Then
Further,
But,
Hence
34.19
Find dy/dx and d
2 y/dx
2 along the general curve x = x(t), y = y(t).
Using dot notation for t-derivatives, we have
34.20 Find the angle at which the cycloid x = a0 - a sin 0, y = a - a cos 0 meets the x-axis at the origin.
dxlde = a - a cos 0, dyldO-asmO. Hence, dy/dx = sin 01(1 - cos 0) = cot 6/2, lim cot (612} = +°°.
Therefore, the cycloid conies in vertically at the origin.
34.21
Find the slope of the curve x = t
5 + sin2irt, y = t + e' at t = l.
= t
4 + 2ircos2irt,
= l + e'. Hence, for t=l,
34.22
Find the slope of the curve x = t
2 +e', y = t+e' at the point (1,1).
dxldt = 2t + e', dyldt =1 + e'. Hence,
The point (1,1) corresponds to the parameter
value t = 0. So, the slope is dy/dx = \=2.
34.23
Find dy/dx and d
2 y/dx
2 for x = a cos
3 0, y = a sin
3 6.
dxlde = -3a cos
2 0 sin 0.
dyldd = 3a sin
2 0 cos 0.
So,
Further,
= —sec
2 6, and
34.24
Find the slope of x - e ' cos 2t, y = e
2t sin 2t at t = 0.
So,
dxldt = -2e ' sin It - e ' cos 2t, dyldt = 2e
2l cos 2t - 2e
2t sin 2t. At f = 0, dxldt = -1, dyldt = 2.
= -2.
34.25
Find the coordinates of the highest point of the curve x = 96t, y = 96t - 16t
2 .
We must maximize y. dy/dt = 96-32t, d
2 yldt
2 = -32. So, the only critical number is t = 3, and, by
the second-derivative test, we have a relative (and, therefore, an absolute) maximum. When t = 3, x = 288,
y = 144.
34.26
Find an equation of the tangent line to the curve x = 3e', y = 5e ' at t = 0.
dxldt = 3e',
dyldt = -5e~',
At t = 0, dy/dx = -1, x = 3, y = 5. Hence,
the tangent line is y - 5 = - § (x - 3), 3y-l5 = -5x + 15, 5x + 3y - 30 = 0. Another method. At / = 0,
the tangent vector (dxldt, dyldt) = (3, -5), so the normal vector is (5, 3). Then, by Problem 33.6, the tangent
line is given by 5jc + 3y + c = 0, where c is determined by the condition that the point (x, y) l=0 = (3,5)
lies on the line.
277
34.18
Find dy/dx and d
2 y/dx
2 for x = t* + t, y = i + t+\.
dx/dt = 3t
2 + l, dy/dt = 7t
6 + l. Then
Further,
But,
Hence
34.19
Find dy/dx and d
2 y/dx
2 along the general curve x = x(t), y = y(t).
Using dot notation for t-derivatives, we have
34.20 Find the angle at which the cycloid x = a0 - a sin 0, y = a - a cos 0 meets the x-axis at the origin.
dxlde = a - a cos 0, dyldO-asmO. Hence, dy/dx = sin 01(1 - cos 0) = cot 6/2, lim cot (612} = +°°.
Therefore, the cycloid conies in vertically at the origin.
34.21
Find the slope of the curve x = t
5 + sin2irt, y = t + e' at t = l.
= t
4 + 2ircos2irt,
= l + e'. Hence, for t=l,
34.22
Find the slope of the curve x = t
2 +e', y = t+e' at the point (1,1).
dxldt = 2t + e', dyldt =1 + e'. Hence,
The point (1,1) corresponds to the parameter
value t = 0. So, the slope is dy/dx = \=2.
34.23
Find dy/dx and d
2 y/dx
2 for x = a cos
3 0, y = a sin
3 6.
dxlde = -3a cos
2 0 sin 0.
dyldd = 3a sin
2 0 cos 0.
So,
Further,
= —sec
2 6, and
34.24
Find the slope of x - e ' cos 2t, y = e
2t sin 2t at t = 0.
So,
dxldt = -2e ' sin It - e ' cos 2t, dyldt = 2e
2l cos 2t - 2e
2t sin 2t. At f = 0, dxldt = -1, dyldt = 2.
= -2.
34.25
Find the coordinates of the highest point of the curve x = 96t, y = 96t - 16t
2 .
We must maximize y. dy/dt = 96-32t, d
2 yldt
2 = -32. So, the only critical number is t = 3, and, by
the second-derivative test, we have a relative (and, therefore, an absolute) maximum. When t = 3, x = 288,
y = 144.
34.26
Find an equation of the tangent line to the curve x = 3e', y = 5e ' at t = 0.
dxldt = 3e',
dyldt = -5e~',
At t = 0, dy/dx = -1, x = 3, y = 5. Hence,
the tangent line is y - 5 = - § (x - 3), 3y-l5 = -5x + 15, 5x + 3y - 30 = 0. Another method. At / = 0,
the tangent vector (dxldt, dyldt) = (3, -5), so the normal vector is (5, 3). Then, by Problem 33.6, the tangent
line is given by 5jc + 3y + c = 0, where c is determined by the condition that the point (x, y) l=0 = (3,5)
lies on the line.